Frequency and Period of SHM - College Physics

Two masses undergo simple harmonic motion with the same amplitude. Mass 1 has twice the frequency of mass 2. What is the ratio of the maximum kinetic energy of mass 1 to the maximum kinetic energy of mass 2, given that mass 1 has half the mass of mass 2?
Answer: 2:1 The maximum kinetic energy in SHM equals the total energy: $KE_{max} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$ . Given $m_1 = m_2/2$ , $f_1 = 2f_2$ (so $\omega_1 = 2\omega_2$ ), and $A_1 = A_2 = A$ . Therefore: $KE_{1,max} = \frac{1}{2}m_1\omega_1^2A^2 = \frac{1}{2}(m_2/2)(2\omega_2)^2A^2 = \frac{1}{2}(m_2/2)(4\omega_2^2)A^2 = m_2\omega_2^2A^2$ and $KE_{2,max} = \frac{1}{2}m_2\omega_2^2A^2$ . The ratio is $\frac{KE_{1,max}}{KE_{2,max}} = \frac{m_2\omega_2^2A^2}{\frac{1}{2}m_2\omega_2^2A^2} = 2$ . A assumes effects cancel. B inverts the ratio. D only considers frequency effect without mass correction.
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?
Answer: $2\pi\sqrt{\frac{0.1}{9.8}}$ 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.
A horizontal mass-spring system has a mass of 0.5 kg and oscillates with a frequency of 2.0 Hz. What is the spring constant of the spring?
Answer: 79 N/m When you encounter oscillation problems involving mass and frequency, you're dealing with simple harmonic motion. The key relationship here connects the natural frequency of a mass-spring system to its physical properties through the equation $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ , where f is frequency, k is the spring constant, and m is the mass. To find the spring constant, rearrange this equation by solving for k. First, multiply both sides by $2\pi$ : $2\pi f = \sqrt{\frac{k}{m}}$ . Then square both sides: $(2\pi f)^2 = \frac{k}{m}$ . Finally, multiply by m: $k = m(2\pi f)^2$ . Substituting the given values: $k = (0.5 \text{ kg})(2\pi \times 2.0 \text{ Hz})^2 = (0.5)(4\pi)^2 = (0.5)(16\pi^2) = 8\pi^2 \approx 79 \text{ N/m}$ . This confirms answer B. Looking at the wrong answers: A (39 N/m) results from forgetting to square the $2\pi f$ term—a common algebraic error when rearranging the frequency equation. C (158 N/m) comes from using the wrong mass value, perhaps doubling it to 1.0 kg instead of 0.5 kg. D (316 N/m) occurs when you forget the factor of $\frac{1}{2\pi}$ in the original frequency equation, essentially using $f = \sqrt{\frac{k}{m}}$ instead. Remember: always double-check your algebra when rearranging the frequency equation for simple harmonic motion. The $2\pi$ factor is crucial and easily mishandled during calculations.
A torsional oscillator consists of a disk suspended by a wire. When the disk has moment of inertia $I_0$ , the period is $T_0 = 1.2$ s. A second identical disk is then attached concentrically to the first, doubling the total moment of inertia. What is the new period?
Answer: 1.7 s When you encounter torsional oscillator problems, focus on the relationship between period and moment of inertia. A torsional oscillator follows the equation $T = 2\pi\sqrt{\frac{I}{\kappa}}$ , where $T$ is the period, $I$ is the moment of inertia, and $\kappa$ is the torsional spring constant of the wire. The key insight is that period is proportional to the square root of moment of inertia: $T \propto \sqrt{I}$ . When the second disk doubles the moment of inertia from $I_0$ to $2I_0$ , the new period becomes: $\frac{T_{new}}{T_0} = \sqrt{\frac{2I_0}{I_0}} = \sqrt{2} \approx 1.41$ Therefore: $T_{new} = 1.2 \text{ s} \times 1.41 = 1.7 \text{ s}$ Answer D (1.7 s) is correct. Answer A (0.85 s) incorrectly assumes the period decreases with increased inertia, perhaps confusing this with the inverse relationship in some other oscillatory systems. Answer B (1.2 s) wrongly suggests the period remains unchanged, ignoring the effect of doubling the moment of inertia entirely. Answer C (2.4 s) represents the trap of assuming a linear relationship—doubling inertia doubles the period—rather than the correct square root relationship. Remember: in oscillatory motion problems, always check whether the relationship is linear, inverse, or involves square roots. Torsional and simple pendulum periods both involve square root relationships with their restoring parameters, while springs follow different patterns.
A vertical spring-mass system oscillates with period $T = 2.0$ s. If the same spring and mass are oriented horizontally on a frictionless surface, what will be the new period?
Answer: 2.0 s The period of a mass-spring system is $T = 2\pi\sqrt{m/k}$ , which depends only on the mass and spring constant, not on the orientation or presence of gravity. When vertical, the spring has a new equilibrium position due to the gravitational force, but this doesn't change the period of small oscillations about that equilibrium. The restoring force is still $F = -kx$ where $x$ is displacement from equilibrium. Therefore, the period remains 2.0 s. A and C assume gravity affects the period. D reflects confusion about what parameters determine the period.
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?
Answer: The amplitude will be significantly larger than the driving amplitude 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.
A spring-mass system undergoes 20 complete oscillations in 8.0 seconds. What is the frequency of the oscillations?
Answer: 2.5 Hz 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.
A mass-spring system oscillates with angular frequency $\omega = 4.0$ rad/s. What is the period of oscillation?
Answer: 1.57 s 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.
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?
Answer: 1.0 s 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).
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?
Answer: $f/2$ 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.
A mass $m$ undergoes simple harmonic motion with period $T_0$ . If the mass is increased to $4m$ and the spring constant is decreased to $k/4$ , what is the new period?
Answer: $4T_0$ The period of SHM is $T = 2\pi\sqrt{\frac{m}{k}}$ . Initially, $T_0 = 2\pi\sqrt{\frac{m}{k}}$ . After the changes, $T_{new} = 2\pi\sqrt{\frac{4m}{k/4}} = 2\pi\sqrt{\frac{16m}{k}} = 2\pi\sqrt{16}\sqrt{\frac{m}{k}} = 4 \cdot 2\pi\sqrt{\frac{m}{k}} = 4T_0$ . A assumes the changes cancel out completely. B only accounts for the mass change or spring constant change, not both. D multiplies the individual effects: $2 \times 2 \times 2 = 8$ .
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?
Answer: 3.0 s 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."
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?
Answer: 4.9 s 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.
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?
Answer: The period would decrease by a factor of $\sqrt{2}$ 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.
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 $k_A$ ?
Answer: 4 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.
A pendulum clock keeps accurate time when the temperature is 20°C. The pendulum rod is made of aluminum with coefficient of linear expansion $\alpha = 23 \times 10^{-6}$ /°C. If the temperature increases to 35°C, by what percentage will the clock run slow?
Answer: 0.017% The length changes by $\Delta L = L_0\alpha\Delta T = L_0(23 \times 10^{-6})(15) = 3.45 \times 10^{-4}L_0$ . The new length is $L = L_0(1 + 3.45 \times 10^{-4})$ . Since $T = 2\pi\sqrt{L/g}$ , the new period is $T = T_0\sqrt{1 + 3.45 \times 10^{-4}} \approx T_0(1 + 1.725 \times 10^{-4})$ using the approximation $\sqrt{1+x} \approx 1 + x/2$ for small $x$ . The clock runs slow by $1.725 \times 10^{-4} = 0.017\%$ . B uses the full thermal expansion without the square root approximation. C and D involve calculation errors or wrong temperature differences.
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?
Answer: $T_B = T_A$ 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.
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?
Answer: 3T 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.
A mass on a spring oscillates horizontally with amplitude $A$ . At what displacement from equilibrium is the speed equal to half the maximum speed?
Answer: $\frac{A\sqrt{3}}{2}$ When analyzing simple harmonic motion, you need to understand the relationship between position, velocity, and energy. For a mass on a spring, the total mechanical energy remains constant and equals the maximum kinetic energy (which occurs at equilibrium) or the maximum potential energy (which occurs at maximum displacement). The velocity at any position $x$ is given by $v = \omega\sqrt{A^2 - x^2}$ , where $\omega$ is the angular frequency and $A$ is the amplitude. The maximum velocity occurs at equilibrium ( $x = 0$ ): $v_{max} = \omega A$ . To find where the speed equals half the maximum speed, set $v = \frac{v_{max}}{2}$ : $\omega\sqrt{A^2 - x^2} = \frac{\omega A}{2}$ Dividing by $\omega$ and squaring both sides: $A^2 - x^2 = \frac{A^2}{4}$ Solving for $x^2$ : $x^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$ Therefore: $x = \frac{A\sqrt{3}}{2}$ This confirms answer (A) is correct. Answer (B) $\frac{A}{2}$ would give you a speed of $\frac{\sqrt{3}}{2}$ times the maximum speed, not half. Answer (C) $\frac{A}{4}$ would give approximately 0.97 times the maximum speed. Answer (D) $\frac{3A}{4}$ would give approximately 0.66 times the maximum speed. Remember: Energy conservation problems in oscillatory motion often involve the relationship $v^2 + \omega^2x^2 = \omega^2A^2$ . This fundamental equation connects position and velocity at any point in the oscillation.
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?
Answer: 0.67 Hz 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.

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