College Physics Quiz: Frequency And Period Of Shm

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Frequency And Period Of ShmQuestion 1 of 20

A mass attached to a spring oscillates with simple harmonic motion. If the spring constant is doubled while keeping the mass constant, how does the frequency of oscillation change?

The frequency increases by a factor of

\sqrt{2}

The frequency increases by a factor of 2

The frequency decreases by a factor of

\sqrt{2}

The frequency remains the same

The frequency decreases by a factor of 2

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College Physics Quiz

College Physics Quiz: Frequency And Period Of Shm

Practice Frequency And Period Of Shm in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Frequency And Period Of Shm, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A mass attached to a spring oscillates with simple harmonic motion. If the spring constant is doubled while keeping the mass constant, how does the frequency of oscillation change?

The frequency increases by a factor of $\sqrt{2}$ (correct answer)
The frequency increases by a factor of 2
The frequency decreases by a factor of $\sqrt{2}$
The frequency remains the same
The frequency decreases by a factor of 2

Explanation: When you encounter simple harmonic motion problems, focus on the fundamental relationship between the physical parameters and the resulting motion. The key equation for a mass-spring system is the frequency formula:

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

, where

k

is the spring constant and

m

is the mass. To find how frequency changes when the spring constant doubles, substitute

2k

for

k

f_{new} = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} = \frac{1}{2\pi}\sqrt{2}\sqrt{\frac{k}{m}} = \sqrt{2} \cdot f_{original}

. Therefore, the frequency increases by a factor of

\sqrt{2}

, making choice A correct. Choice B suggests the frequency doubles, which would occur if frequency were directly proportional to spring constant rather than proportional to its square root. This is a common misconception—students often forget about the square root relationship. Choice C claims the frequency decreases by

\sqrt{2}

, which would happen if you mistakenly inverted the relationship, thinking stiffer springs oscillate more slowly. Choice D suggests no change in frequency, ignoring the spring constant's role entirely in the motion. Remember that in simple harmonic motion, frequency depends on the square root of the restoring force parameter (like spring constant) divided by the inertial parameter (like mass). This

\sqrt{2}

factor appears frequently in physics when you double a quantity that's under a square root, so watch for this pattern in oscillation, wave, and energy problems.

Question 2

A pendulum has a period of 2.0 s on Earth. What would be its period on a planet where the gravitational acceleration is 4 times that of Earth?

0.5 s
1.0 s (correct answer)
2.0 s
4.0 s
8.0 s

Explanation: When you encounter pendulum problems involving different gravitational environments, you need to understand how the period depends on gravitational acceleration through the pendulum equation. The period of a simple pendulum is given by

T = 2\pi\sqrt{\frac{L}{g}}

, where L is the length and g is gravitational acceleration. Notice that period is inversely proportional to the square root of g. This means if gravity increases, the period decreases. Starting with Earth's conditions:

T_{Earth} = 2\pi\sqrt{\frac{L}{g_{Earth}}} = 2.0 \text{ s}

On the new planet where

g_{planet} = 4g_{Earth}

T_{planet} = 2\pi\sqrt{\frac{L}{4g_{Earth}}} = 2\pi\sqrt{\frac{L}{g_{Earth}}} \cdot \sqrt{\frac{1}{4}} = T_{Earth} \cdot \frac{1}{2} = 1.0 \text{ s}

So the correct answer is B) 1.0 s. Let's examine why the other options are wrong: A) 0.5 s would result if you mistakenly used the full factor of 4 instead of its square root. C) 2.0 s suggests gravity has no effect on pendulum motion, which ignores the fundamental relationship. D) 4.0 s implies the period increases with stronger gravity, which is backwards—you might get this by multiplying instead of dividing. Study tip: Remember that pendulum period scales with

\frac{1}{\sqrt{g}}

. When gravity changes by a factor, take the square root and flip it to find how period changes. Stronger gravity always makes pendulums swing faster (shorter periods).

Question 3

A simple pendulum of length L has a period T. If the length is increased to 9L, what is the new period in terms of T?

T/3
T/9
3T (correct answer)
9T
81T

Explanation: When you encounter pendulum problems, focus on the relationship between period and length. The period of a simple pendulum depends only on its length and gravitational acceleration, following the formula

T = 2\pi\sqrt{\frac{L}{g}}

. To find how the period changes when length increases from L to 9L, you can set up a ratio. The new period is

T_{new} = 2\pi\sqrt{\frac{9L}{g}}

. Since

\sqrt{9L} = \sqrt{9} \times \sqrt{L} = 3\sqrt{L}

, the new period becomes

T_{new} = 2\pi \times 3\sqrt{\frac{L}{g}} = 3 \times 2\pi\sqrt{\frac{L}{g}} = 3T

. The period triples when the length increases by a factor of 9. Looking at the wrong answers: Choice A (T/3) represents the common error of thinking period is inversely proportional to length, perhaps confusing this with frequency relationships. Choice B (T/9) suggests a direct inverse relationship with the length multiplier, ignoring the square root in the formula. Choice D (9T) assumes period is directly proportional to length without considering the square root relationship. Remember that for pendulum problems, period is proportional to the square root of length. When length changes by a factor of n, period changes by a factor of

\sqrt{n}

. This square root relationship is crucial – it means doubling the length doesn't double the period, and increasing length by 9 times increases period by only 3 times.

Question 4

Two pendulums have the same length but different masses: $m_1 = 2.0$ kg and $m_2 = 8.0$ kg. How do their periods compare?

$T_2 = 4T_1$
$T_2 = 2T_1$
$T_2 = T_1$ (correct answer)
$T_2 = T_1/2$
$T_2 = T_1/4$

Explanation: When you encounter pendulum problems, the key insight is understanding what factors actually affect the period of oscillation. This is a classic physics concept that often surprises students. For a simple pendulum, the period is given by

T = 2\pi\sqrt{\frac{L}{g}}

, where

L

is the length and

g

is gravitational acceleration. Notice what's missing from this formula: mass. The period of a pendulum depends only on its length and the strength of gravity, not on how heavy the bob is. Since both pendulums have the same length, they must have identical periods:

T_2 = T_1

. This makes C correct. Let's examine why the other answers reflect common misconceptions. Answer A (

T_2 = 4T_1

) suggests the period is proportional to mass squared, since

8.0/2.0 = 4

. Answer B (

T_2 = 2T_1

) implies the period is directly proportional to mass, again using the mass ratio

8.0/2.0 = 2

. Answer D (

T_2 = T_1/2

) suggests an inverse relationship with mass. All of these stem from the intuitive but incorrect assumption that heavier objects should swing differently. The physics behind this counterintuitive result is that while a heavier mass experiences more gravitational force, it also has more inertia to overcome. These effects exactly cancel out, leaving period independent of mass. Remember: For simple pendulums, only length matters for the period. Mass independence is a fundamental characteristic that distinguishes pendulum motion from many other oscillatory systems.

Question 5

A physical pendulum consists of a uniform rod of length L pivoted at one end. If the rod were cut to half its length (L/2), how would the period change?

The period would decrease by a factor of 2
The period would decrease by a factor of $\sqrt{2}$ (correct answer)
The period would remain the same
The period would increase by a factor of $\sqrt{2}$
The period would increase by a factor of 2

Explanation: When analyzing physical pendulum problems, you need to understand how the period depends on both the moment of inertia and the distance to the center of mass. The period formula for a physical pendulum is

T = 2\pi\sqrt{\frac{I}{mgd}}

, where I is the moment of inertia about the pivot, m is mass, and d is the distance from pivot to center of mass. For a uniform rod of length L pivoted at one end, the moment of inertia is

I = \frac{1}{3}mL^2

and the center of mass is at

d = \frac{L}{2}

. This gives

T_{original} = 2\pi\sqrt{\frac{mL^2/3}{mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}

. When you cut the rod to length L/2, both the moment of inertia and center of mass distance change. The new moment of inertia becomes

I_{new} = \frac{1}{3}m_{new}(L/2)^2 = \frac{1}{12}mL^2

(assuming the mass also halves), and the new center of mass distance is

d_{new} = L/4

. The new period is

T_{new} = 2\pi\sqrt{\frac{mL^2/12}{(m/2)g(L/4)} = 2\pi\sqrt{\frac{L}{3g}}

. Comparing the periods:

\frac{T_{new}}{T_{original}} = \sqrt{\frac{L/3g}{2L/3g}} = \frac{1}{\sqrt{2}}

. Therefore, the period decreases by a factor of

\sqrt{2}

, making B correct. Choice A suggests a factor of 2, which would occur if only length mattered linearly. Choice C incorrectly assumes no change. Choice D has the right factor but wrong direction. Remember: physical pendulum problems require considering both rotational inertia and gravitational restoring torque—both change when dimensions change.

Question 6

A mass-spring system oscillates with angular frequency $\omega = 4.0$ rad/s. What is the period of oscillation?

0.25 s
0.64 s
1.57 s (correct answer)
4.0 s
25.1 s

Explanation: This question tests your understanding of the fundamental relationship between angular frequency and period in oscillatory motion. When you encounter any oscillation problem, always identify which quantities you're given and which relationships connect them. The key relationship here is

T = \frac{2\pi}{\omega}

, where

T

is the period and

\omega

is the angular frequency. Given

\omega = 4.0

rad/s, you can substitute directly:

T = \frac{2\pi}{4.0} = \frac{6.283}{4.0} = 1.57 \text{ s}

This confirms answer C is correct. Let's examine why the other options are wrong. Answer A (0.25 s) represents the trap of calculating

T = \frac{1}{\omega}

instead of

T = \frac{2\pi}{\omega}

. This mistake comes from confusing the relationship between period and angular frequency with the simpler relationship between period and regular frequency (

f

), where

T = \frac{1}{f}

. Answer B (0.64 s) might result from computational errors or using an incorrect approximation for

\pi

. Answer D (4.0 s) is simply the given angular frequency value, which students might choose if they confuse period with angular frequency or forget to perform any calculation at all. Remember this key distinction: angular frequency

\omega

(in rad/s) relates to period by

T = \frac{2\pi}{\omega}

, while regular frequency

f

(in Hz) relates to period by

T = \frac{1}{f}

. The

2\pi

factor is crucial when working with angular quantities in oscillatory motion.

Question 7

A mass oscillating on a spring completes one full cycle in 0.8 s. How many complete oscillations will it make in 10 seconds?

8.0
10.8
12.5 (correct answer)
80
125

Explanation: When you encounter oscillation problems, you're dealing with periodic motion where the key relationship is between period (time for one complete cycle) and frequency (cycles per unit time). These are reciprocals of each other. Given that one complete cycle takes 0.8 seconds, you can find how many cycles occur in 10 seconds using a simple proportion. If 1 cycle takes 0.8 s, then in 10 seconds you'll have:

\text{Number of cycles} = \frac{10 \text{ s}}{0.8 \text{ s/cycle}} = 12.5 \text{ cycles}

This confirms answer C is correct. Let's examine why the other options are wrong. Answer A (8.0) comes from incorrectly dividing 0.8 by 10 instead of 10 by 0.8 – this reverses the relationship and gives you a fraction rather than the actual number of cycles. Answer B (10.8) appears to come from adding 10 + 0.8, which has no physical meaning in this context. Answer D (80) results from multiplying 10 × 8, possibly from first calculating 10/0.8 = 12.5, then mistakenly thinking 8 complete cycles means 8 × 10 = 80. Remember that in oscillation problems, you're often converting between period and frequency. The number of complete oscillations equals the total time divided by the period of one oscillation. Always check that your units work out correctly – you should get a dimensionless number when counting cycles.

Question 8

Two mass-spring systems have identical springs but different masses. System A has mass $m$ and system B has mass $4m$ . If system A oscillates with frequency $f$ , what is the frequency of system B?

$f/4$
$f/2$ (correct answer)
$f$
$2f$
$4f$

Explanation: When you encounter oscillating mass-spring systems, you're dealing with simple harmonic motion, where the key relationship is between mass, spring constant, and frequency. The fundamental equation governing this motion is

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

, where

f

is frequency,

k

is the spring constant, and

m

is mass. Since both systems have identical springs, they share the same spring constant

k

. For system A with mass

m

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

. For system B with mass

4m

f_B = \frac{1}{2\pi}\sqrt{\frac{k}{4m}} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \cdot \frac{1}{\sqrt{4}} = f \cdot \frac{1}{2} = \frac{f}{2}

. Therefore, answer B is correct. Looking at the wrong answers: Choice A (

f/4

) incorrectly assumes frequency is inversely proportional to mass rather than the square root of mass. Choice C (

f

) ignores the mass difference entirely, suggesting frequency depends only on the spring constant. Choice D (

2f

) reverses the relationship, treating frequency as if it increases with mass. Remember that in oscillatory systems, frequency decreases with the square root of mass. When mass increases by a factor of 4, frequency decreases by a factor of

\sqrt{4} = 2

. This inverse square root relationship between mass and frequency is fundamental to understanding all oscillating systems, from pendulums to molecular vibrations.

Question 9

A pendulum clock keeps accurate time on Earth where its period is 2.0 s. On the Moon, where gravitational acceleration is 1/6 that of Earth, what would be the period of the same pendulum?

0.82 s
2.0 s
4.9 s (correct answer)
12 s
49 s

Explanation: When you encounter pendulum problems involving different gravitational environments, the key relationship to remember is that a simple pendulum's period depends on the square root of the gravitational acceleration. The period of a simple pendulum is given by

T = 2\pi\sqrt{\frac{L}{g}}

, where L is the length and g is gravitational acceleration. Since the same pendulum is used on both Earth and Moon, the length L remains constant. This means the period is inversely proportional to the square root of gravity:

T \propto \frac{1}{\sqrt{g}}

. On the Moon, gravity is 1/6 that of Earth, so

g_{Moon} = \frac{g_{Earth}}{6}

. Taking the ratio of periods:

\frac{T_{Moon}}{T_{Earth}} = \sqrt{\frac{g_{Earth}}{g_{Moon}}} = \sqrt{\frac{g_{Earth}}{g_{Earth}/6}} = \sqrt{6} \approx 2.45

Therefore:

T_{Moon} = 2.45 \times 2.0\text{ s} = 4.9\text{ s}

Looking at the wrong answers: A) 0.82 s results from incorrectly thinking the period decreases with weaker gravity, possibly by dividing by

\sqrt{6}

instead of multiplying. B) 2.0 s assumes gravity has no effect on the pendulum period, ignoring the fundamental relationship entirely. D) 12 s comes from multiplying by 6 instead of

\sqrt{6}

, missing the square root relationship in the period formula. Remember: weaker gravity makes pendulums swing more slowly (longer periods), and the relationship involves the square root of the gravitational ratio, not the ratio itself.

Question 10

A torsional pendulum consists of a disk suspended by a wire. If the wire is replaced by one that is twice as stiff (twice the torsion constant), how does the frequency change?

The frequency decreases by a factor of $\sqrt{2}$
The frequency decreases by a factor of 2
The frequency remains the same
The frequency increases by a factor of $\sqrt{2}$ (correct answer)
The frequency increases by a factor of 2

Explanation: When you encounter torsional pendulum problems, you're dealing with simple harmonic motion where the restoring force comes from the twisting of a wire or rod. The key relationship to remember is that frequency depends on both the stiffness of the system and its inertia. For a torsional pendulum, the frequency is given by

f = \frac{1}{2\pi}\sqrt{\frac{\kappa}{I}}

, where

\kappa

is the torsion constant (stiffness) and

I

is the moment of inertia of the disk. This formula shows that frequency is directly proportional to the square root of the torsion constant. When the wire is replaced with one that's twice as stiff,

\kappa

doubles while the moment of inertia

I

stays the same (same disk). Therefore:

f_{new} = \frac{1}{2\pi}\sqrt{\frac{2\kappa}{I}} = \sqrt{2} \cdot \frac{1}{2\pi}\sqrt{\frac{\kappa}{I}} = \sqrt{2} \cdot f_{original}

Looking at the wrong answers: A suggests the frequency decreases by

\sqrt{2}

, which would happen if you mistakenly thought stiffer wires make oscillations slower. B implies a linear relationship (factor of 2), missing the square root dependence. C assumes the torsion constant doesn't affect frequency at all, ignoring the fundamental physics of the restoring torque. The correct answer is D: the frequency increases by a factor of

\sqrt{2}

. Study tip: For any harmonic oscillator, remember that frequency increases with the square root of the stiffness parameter. Stiffer systems oscillate faster, not slower, and the relationship involves square roots, not linear factors.

Question 11

A spring with spring constant $k = 100$ N/m supports a $0.25$ kg mass. The mass is displaced 5.0 cm from equilibrium and released. What is the frequency of the resulting oscillations?

3.18 Hz (correct answer)
10.0 Hz
20.0 Hz
31.8 Hz
400 Hz

Explanation: This problem tests simple harmonic motion, specifically the relationship between a mass-spring system's physical properties and its oscillation frequency. When you see a spring-mass system question, focus on the fundamental frequency formula and ignore any initial displacement information—it doesn't affect the frequency. For a mass-spring system, the frequency is given by

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

, where

k

is the spring constant and

m

is the mass. Substituting the given values:

k = 100

N/m and

m = 0.25

kg.

f = \frac{1}{2\pi}\sqrt{\frac{100}{0.25}} = \frac{1}{2\pi}\sqrt{400} = \frac{20}{2\pi} = \frac{10}{\pi} \approx 3.18 \text{ Hz}

Answer A (3.18 Hz) is correct. Answer B (10.0 Hz) represents the common error of forgetting the

2\pi

factor—you'd get this if you calculated

\frac{1}{\pi}\sqrt{\frac{k}{m}}

instead. Answer C (20.0 Hz) comes from omitting the entire

2\pi

denominator and just calculating

\sqrt{\frac{k}{m}}

. Answer D (31.8 Hz) might result from calculation errors or incorrectly manipulating the formula. Remember: The frequency of a spring-mass system depends only on the spring constant and mass—never on the amplitude of oscillation. Always double-check that you're using the complete frequency formula with the

2\pi

factor, as this is the most common source of error in oscillation problems.

Question 12

A mass attached to a spring oscillates vertically. The equilibrium position is 10 cm below the natural length of the spring. What is the period of small oscillations about this equilibrium position?

$2\pi\sqrt{\frac{0.1}{9.8}}$ (correct answer)
$2\pi\sqrt{\frac{mg}{k}}$
$2\pi\sqrt{\frac{m}{k}}$
$2\pi\sqrt{\frac{k}{mg}}$
$2\pi\sqrt{\frac{k}{m}}$

Explanation: When analyzing spring-mass oscillations in a gravitational field, the key insight is that the equilibrium position automatically accounts for gravitational effects, simplifying the physics considerably. At equilibrium, the spring force balances the gravitational force:

kx_0 = mg

, where

x_0 = 0.1

m is the extension from natural length. This gives us

k = \frac{mg}{0.1} = \frac{mg}{0.1}

. For small oscillations about this equilibrium position, gravity doesn't affect the period because we're measuring displacements from where forces already balance. The restoring force for a small displacement

x

from equilibrium is simply

F = -kx

, giving us simple harmonic motion with period

T = 2\pi\sqrt{\frac{m}{k}}

. Substituting our expression for

k

T = 2\pi\sqrt{\frac{m}{\frac{mg}{0.1}}} = 2\pi\sqrt{\frac{0.1}{g}} = 2\pi\sqrt{\frac{0.1}{9.8}}

. This confirms answer A is correct. Answer B,

2\pi\sqrt{\frac{mg}{k}}

, incorrectly treats

mg

as the "effective mass" in the oscillation formula. Answer C,

2\pi\sqrt{\frac{m}{k}}

, is the generic spring-mass formula but fails to use the given equilibrium information to find the actual value of

k

. Answer D,

2\pi\sqrt{\frac{k}{mg}}

, inverts the mass and spring constant relationship entirely. Remember: when a spring-mass system reaches equilibrium under gravity, use that equilibrium condition to relate

k

m

, and

g

. The resulting period will depend on the equilibrium displacement and gravitational acceleration, not abstract spring constants.

Question 13

Two identical pendulums are set up side by side. Pendulum A is started with a small amplitude, and pendulum B is started with twice that amplitude. How do their periods compare?

$T_B = 2T_A$
$T_B = \sqrt{2}T_A$
$T_B = T_A$ (correct answer)
$T_B = T_A/\sqrt{2}$
$T_B = T_A/2$

Explanation: When you encounter pendulum problems, the key insight is understanding what factors actually affect the period. The period of a simple pendulum depends only on the length of the pendulum and gravitational acceleration, not on the amplitude of oscillation. For small amplitudes (typically less than 15°), the period formula is

T = 2\pi\sqrt{\frac{L}{g}}

, where L is the length and g is gravitational acceleration. Notice that amplitude doesn't appear in this equation at all. Since both pendulums are identical (same length) and operating under the same gravity, they must have the same period regardless of their starting amplitudes. This means

T_B = T_A

, making answer C correct. Let's examine why the other options are wrong. Answer A suggests

T_B = 2T_A

, which would imply the period doubles with double amplitude - this confuses amplitude with period. Answer B proposes

T_B = \sqrt{2}T_A

, which might seem reasonable since the amplitude relationship is

\sqrt{2}

, but this incorrectly assumes amplitude affects period. Answer D suggests

T_B = T_A/\sqrt{2}

, which would mean larger amplitude creates shorter periods - also physically incorrect. The key study tip: Remember that for simple pendulums at small amplitudes, period depends only on length and gravity. Amplitude independence is a crucial characteristic that distinguishes simple harmonic motion from other types of oscillation. This concept appears frequently on physics exams, so memorize that identical pendulums always have identical periods regardless of how you start them swinging.

Question 14

A simple pendulum has a period of 3.0 s. If the pendulum bob is replaced with one having twice the mass but the length remains constant, what is the new period?

1.5 s
2.1 s
3.0 s (correct answer)
4.2 s
6.0 s

Explanation: When you encounter pendulum problems, the key insight is understanding what variables actually affect the period. This tests a fundamental principle that often surprises students. The period of a simple pendulum is given by

T = 2\pi\sqrt{\frac{L}{g}}

, where L is the length and g is gravitational acceleration. Notice what's missing from this formula: mass. The period depends only on the length of the pendulum and gravity, not on the mass of the bob. Since the length remains constant when you replace the bob, the period stays exactly the same at 3.0 s. This might seem counterintuitive—you'd expect a heavier mass to swing differently—but gravity affects all objects equally regardless of mass, just like how a feather and hammer fall at the same rate in a vacuum. Looking at the wrong answers: A) 1.5 s represents the misconception that doubling mass halves the period (inverse relationship). B) 2.1 s suggests students might think

T \propto \frac{1}{\sqrt{m}}

, giving

\frac{3.0}{\sqrt{2}} \approx 2.1

s. D) 4.2 s comes from incorrectly assuming

T \propto \sqrt{m}

, yielding

3.0 \times \sqrt{2} \approx 4.2

s. These all reflect the common error of thinking mass affects pendulum motion. Study tip: Memorize that pendulum period depends only on length and gravity:

T = 2\pi\sqrt{\frac{L}{g}}

. Mass independence is a favorite exam trap—when you see mass changes in pendulum problems, immediately think "no effect on period."

Question 15

A mass-spring system oscillates with a period of 4.0 s. If both the mass and spring constant are doubled simultaneously, what is the new period?

1.0 s
2.0 s
4.0 s (correct answer)
8.0 s
16.0 s

Explanation: When you encounter oscillation problems involving changes to system parameters, you need to understand how the period formula responds to simultaneous modifications. For a mass-spring system, the period is given by

T = 2\pi\sqrt{\frac{m}{k}}

, where m is the mass and k is the spring constant. To find the new period when both parameters change, you substitute the new values into this formula. If both mass and spring constant are doubled, the new period becomes:

T_{new} = 2\pi\sqrt{\frac{2m}{2k}} = 2\pi\sqrt{\frac{m}{k}} = T_{original}

The factors of 2 in the numerator and denominator cancel out completely, leaving the period unchanged at 4.0 s. Looking at the wrong answers: Choice A (1.0 s) represents the mistake of thinking both doubling effects work to decrease the period, perhaps by dividing the original period by 4. Choice B (2.0 s) comes from incorrectly assuming that doubling the spring constant alone dominates, since a stiffer spring does decrease period by a factor of √2. Choice D (8.0 s) results from thinking that doubling the mass alone dominates, since more mass does increase the period. The key insight is recognizing that mass and spring constant have opposite effects on period—mass increases it while spring constant decreases it. When both are changed by the same factor, these effects exactly cancel out. Remember this cancellation principle: equal multiplicative changes to parameters that appear in a ratio leave the result unchanged.

Question 16

A spring-mass system undergoes 20 complete oscillations in 8.0 seconds. What is the frequency of the oscillations?

0.4 Hz
2.5 Hz (correct answer)
8.0 Hz
20 Hz
160 Hz

Explanation: When you encounter oscillation problems, you're working with periodic motion where frequency measures how many complete cycles occur per unit time. The key is distinguishing between the total number of oscillations and the time period over which they occur. To find frequency, you need to apply the fundamental definition: frequency equals the number of complete oscillations divided by the total time. Here, the spring-mass system completes 20 oscillations in 8.0 seconds, so:

f = \frac{\text{number of oscillations}}{\text{time}} = \frac{20 \text{ oscillations}}{8.0 \text{ seconds}} = 2.5 \text{ Hz}

This confirms answer choice B is correct. Looking at the wrong answers: Choice A (0.4 Hz) results from incorrectly dividing time by oscillations (8.0/20), which gives you period instead of frequency. Choice C (8.0 Hz) mistakenly uses the time value as the frequency, ignoring the oscillation count entirely. Choice D (20 Hz) incorrectly treats the number of oscillations as the frequency, completely overlooking the time factor. The key insight is that frequency and period are reciprocals: if you accidentally calculate one, you can easily find the other. Notice that 0.4 Hz × 2.5 Hz = 1, confirming this reciprocal relationship between choices A and B. Study tip: Always write down the frequency formula

f = \frac{N}{t}

first, then identify which values correspond to N (number of cycles) and t (time). This prevents mixing up the calculation and helps you avoid the common trap of calculating period instead of frequency.

Question 17

A mass attached to a vertical spring oscillates with simple harmonic motion. The system has a period of 1.5 s. What is the frequency of oscillation?

0.33 Hz
0.67 Hz (correct answer)
1.5 Hz
3.0 Hz
9.4 Hz

Explanation: When you encounter simple harmonic motion problems, you're dealing with the fundamental relationship between period and frequency. These two quantities describe the same oscillatory motion from different perspectives: period tells you how long one complete cycle takes, while frequency tells you how many cycles occur per unit time. The relationship between period (T) and frequency (f) is straightforward:

f = \frac{1}{T}

. Since the spring-mass system has a period of 1.5 s, the frequency is

f = \frac{1}{1.5 \text{ s}} = 0.67 \text{ Hz}

. This means the mass completes about two-thirds of an oscillation cycle every second. Looking at the wrong answers: Choice A (0.33 Hz) represents taking the reciprocal incorrectly—perhaps confusing the calculation or making an arithmetic error. Choice C (1.5 Hz) is the classic trap of confusing period and frequency; students sometimes think these quantities are equal rather than reciprocals. Choice D (3.0 Hz) might result from multiplying instead of dividing, showing a fundamental misunderstanding of the period-frequency relationship. The key insight is that period and frequency are always reciprocals in any periodic motion. A longer period means lower frequency (fewer cycles per second), while a shorter period means higher frequency (more cycles per second). Remember this inverse relationship: as one increases, the other decreases proportionally. Always double-check that your frequency calculation gives a reasonable result—if the period is greater than 1 second, the frequency must be less than 1 Hz.

Question 18

A mass-spring system has a natural frequency of 5.0 Hz. If the system is driven at a frequency of 5.0 Hz with small amplitude, what characterizes the steady-state response?

The amplitude will be zero due to destructive interference
The amplitude will be the same as the driving amplitude
The amplitude will be significantly larger than the driving amplitude (correct answer)
The frequency of oscillation will be twice the driving frequency
The system will oscillate at its natural frequency, ignoring the driving force

Explanation: This question tests your understanding of resonance in driven oscillatory systems. When you encounter problems about driving frequencies matching natural frequencies, you're dealing with the phenomenon of resonance. When a mass-spring system is driven at exactly its natural frequency (5.0 Hz in this case), resonance occurs. At resonance, the driving force is perfectly synchronized with the system's natural motion, continuously adding energy to the system. Even with a small driving amplitude, this continuous energy input causes the oscillation amplitude to grow significantly larger than the original driving amplitude. This amplification effect is the hallmark of resonance and makes option C correct. Let's examine why the other answers are incorrect. Option A suggests destructive interference would cause zero amplitude, but this confuses wave interference with mechanical resonance - when frequencies match in driven oscillators, you get constructive energy transfer, not destructive interference. Option B claims the amplitude equals the driving amplitude, which would only be true for off-resonance driving where no amplification occurs. Option D proposes frequency doubling, but in linear systems, the steady-state response always oscillates at the driving frequency, not some multiple of it. The key insight is that resonance amplifies small inputs into large outputs. Remember this pattern: when driving frequency equals natural frequency, expect dramatic amplitude amplification. This principle applies across physics, from mechanical systems to electrical circuits to architectural design considerations.

Question 19

Two identical masses are attached to springs with different spring constants. Mass A oscillates with frequency $f_A = 3.0$ Hz, and mass B oscillates with frequency $f_B = 6.0$ Hz. What is the ratio of spring constant $k_B$ to spring constant $k_A$ ?

2
3
4 (correct answer)
6
12

Explanation: When you encounter oscillation problems involving springs and masses, you're dealing with simple harmonic motion. The key relationship to remember is that the frequency of oscillation depends on both the mass and the spring constant according to

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

. Since both masses are identical, you can set up a ratio to find the relationship between the spring constants. For mass A:

f_A = \frac{1}{2\pi}\sqrt{\frac{k_A}{m}}

and for mass B:

f_B = \frac{1}{2\pi}\sqrt{\frac{k_B}{m}}

. Taking the ratio

\frac{f_B}{f_A} = \sqrt{\frac{k_B}{k_A}}

, you get

\frac{6.0}{3.0} = 2 = \sqrt{\frac{k_B}{k_A}}

. Squaring both sides gives

\frac{k_B}{k_A} = 4

, which is answer C. Let's examine why the other options are incorrect. Choice A (2) represents the ratio of frequencies, not spring constants—this is a common trap where students forget to square the frequency ratio. Choice B (3) has no clear mathematical relationship to the given frequencies and likely comes from incorrect arithmetic. Choice D (6) might result from multiplying the frequencies instead of taking their ratio, showing a fundamental misunderstanding of the relationship. Remember this key insight: frequency is proportional to the square root of the spring constant, so when comparing spring constants, you must square the frequency ratio. This relationship appears frequently in oscillation problems, so mastering it will serve you well on similar questions.

Question 20

Two identical springs with spring constant $k$ are connected to a mass $m$ in different configurations. In configuration A, the springs are in series (end-to-end). In configuration B, the springs are in parallel (side-by-side). What is the ratio of the frequency in configuration B to the frequency in configuration A?

$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$
$2$ (correct answer)

Explanation: For springs in series:

\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

, so

k_{eff} = \frac{k}{2}

. For springs in parallel:

k_{eff} = k + k = 2k

. The frequency is

f = \frac{1}{2\pi}\sqrt{\frac{k_{eff}}{m}}

. Therefore

\frac{f_B}{f_A} = \sqrt{\frac{k_{eff,B}}{k_{eff,A}}} = \sqrt{\frac{2k}{k/2}} = \sqrt{4} = 2

. A inverts the ratio. B assumes the effective spring constants are

k/\sqrt{2}

and

k\sqrt{2}

. C takes the square root of the correct answer.