AP Physics 1 - Period and Frequency of Harmonic Motion

A horizontal spring with a constant $k = 250\ \frac{\textup{N}}{\textup{m}}$ is on a frictionless surface. If the mass is doubled, by what factor is the frequency of the spring changed? Assume simple harmonic motion.

$\sqrt\frac{1}{2}$

$\sqrt{2}$

$\frac{1}{2}$

Explanation

The expression for the frequency of a spring:

$f = \frac{w}{2\pi}$

Therefore, we can say:

$\frac{f_1}{f_2}=\frac{\left ( \frac{w_1}{2\pi}\right )}{\left ( \frac{w_2}{2\pi}\right )}$

$\frac{f_1}{f_2}=\frac{w_1}{w_2}$

Where:

$w = \sqrt{\frac{k}{m}}$

Substituting this into our expression, we get:

$\frac{f_1}{f_2}=\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_2}}}$

$\frac{f_1}{f_2}= \sqrt{\frac{m_2}{m_1}}$

Taking the inverse of both sides:

$\frac{f_2}{f_1}= \sqrt{\frac{m_1}{m_2}}$

Rearranging for final frequency:

$f_2= f_1\sqrt{\frac{m_1}{m_2}}$

From the problem statement, we know that:

substituting this into the expression, we get:

$f_2 = f_1\sqrt{\frac{m_1}{2m_2}} = f_1\sqrt{\frac{1}{2}}$

Therefore, the frequency was changed by a factor of $\sqrt{\frac{1}{2}}$

A simple pendulum has a block of mass $m = 50\textup{ kg}$ attached to one end and is rotating in simple harmonic motion. If the frequency of the pendulum is $0.4\textup{ Hz}$ , what is the length of the pendulum?

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

$1.6\textup{ m}$

$3.8\textup{ m}$

$5.2\textup{ m}$

$7.9\textup{ m}$

$9.1\textup{ m}$

Explanation

We only need one expression to solve this problem:

$f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$

Now we just need to rearrange for the length of the pendulum:

$2\pi f = \sqrt{\frac{g}{L}}$

$\frac{g}{L} = (2\pi f)^2$

$L = \frac{g}{(2\pi f)^2}$

We have values for each of these variables, so time to plug and chug:

$L = \frac{\left ( 10\frac{m}{s^2}\right )}{\left ( 2\pi 0.4\frac{1}{s}\right )^2}$

A horizontal spring on a frictionless surface has a spring constant of $10\frac{N}{m}$ with a mass of attached to the end of the spring. If the spring is stretched passed its point of equilibrium and released, how many times does the mass pass through equilibrium per minute?

Explanation

We need to know three things to solve this problem.

Formula for frequency of a spring in simple harmonic motion
The spring constant
Mass of object attached to spring

We're given 2 and 3, so we just need to know 1.

The formula for frequency is:

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

Plugging in our values, we get:

$f = \frac{1}{2\pi}\sqrt{\frac{10\frac{N}{m}}{2kg}} = \frac{1}{2\pi}\sqrt{5\frac{1}{s^2}}$

Now we need to convert Hertz (cycles per second) to cycles per minute.

To get to these units:

$f = 0.356 \frac{1}{s}(\frac{60s}{min}) = 21.4\ min^{-1}$

The position of a mass in an oscillating spring-mass system is given by the following equation:

$x(t)=2sin(4\pi t)+4$ , where is measured in , and is measured in .

What is the period of the oscillations?

$\frac{1}{2}s$

$\frac{1}{4}s$

Explanation

In trigonometric functions, the period is always given by $\frac{2\pi}{B}$ , when the function is written as . Once, we determine our value, we are halfway to the solution!

$B=4\pi$

$P=\frac{2\pi}{B}=\frac{2\pi}{4\pi}=\frac{1}{2}s$

Determine the period of a sine wave that has a frequency of .

$\frac{1}{2}s$

Explanation

Period is given by: $\frac{1}{f}$ where is frequency. Therefore,

$period=\frac{1}{2Hz}=\frac{1}{2}s$

A block of mass 1.3 kg is attached to a spring whose force constant, , is $250 :\frac{N}{m}$ . What is the frequency and period of the oscillations of this spring-block system?

Explanation

For a mass-spring system undergoing simple harmonic motion, the frequency of the oscillations can be found using the equation

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

We were given the force constant (or spring constant), , to be $250 :\frac{N}{m}$ . The oscillating mass was also given to be 1.3 kg. So, plug these in to the equation and solve for frequency, . The unit for frequency is Hertz, Hz.

$f=\frac{1}{2\pi }\sqrt{\frac{250:\frac{N}{m}}{1.3:kg}}=2.2:Hz$

Find the mass of the bob of a simple pendulum if the period of the pendulum is seconds, and the length of the pendulum is .

Impossible to determine

$5.07*10^{-3}kg$

Explanation

It's impossible because the period of a simple pendulum doesn't depend on the mass of the bob. Because of this, we have no way to determine the mass from the period.

As on object passes through its equilibrium position during simple harmonic motion, which statements are true regarding its potential (U) and kinetic (K) energies?

min U, max K

max U, min K

min U, min K

max U, max K

None of these

Explanation

An object has the maximum potential energy the furthest from its equilibrium point (at the turnaround point). So it at the equilibrium position it would have the minimum potential energy. If it is undergoing simple harmonic motion, it would have the maximum kinetic energy as it passes through the equilibrium position because it is returning from the stretched position (spring example) where it gathered energy. The same is true for other objects undergoing this motion.

Consider the following system:

Pendulum_1

If the mass reaches a maximum height of and the minimum angle is $A = 30^{\circ}$ , what is the period of the pendulum?

$g = 10\frac{m}{s^2}$

Explanation

First, we will be using the equation for the period of a pendulum:

$T = 2\pi \sqrt{\frac{l}{g}}$

The only value we don't have is length. However, we can develop an expression for length from the given information.

$h_{max} = l-lsin(A_{min})$

The second term describes how close the pendulum gets to the height of the top of the pendulum. Therefore, we subtract this value from the lowest point of the pendulum to get the height relative to its lowest point. We can rewrite this as:

$h_{max}= l(1-sin(A_{min}))$

$l = \frac{h_{max}}{1-sin(A_{min})}$

Substituting this into the original expression for the period, we get:

$T = 2\pi\sqrt{ \frac{h_{max}}{g(1-sin(A_{min}))}}$

We can use our given values to solve:

$T = 2\pi \sqrt{ \frac{0.75m}{(10\frac{m}{s^2})(1-sin(30^{\circ}))}}$

A 300 g mass is attached to a spring and undergoes simple harmonic motion with a period of 0.25 s. If the total energy of the system is 3.0 J, what is the aplitude of the oscillations?

Explanation

We were given the mass of the system as 300 g. First, we should convert this to kilograms. Since , we can convert by

$m=300:g\cdot \frac{1:kg}{1000:g}=0.3:kg$

They also told us the period of the oscillations, which is 0.25 s. We can use the following equation to solve for the force constant:

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

Rearrange the equation to isolate the variable , and substitute in the known values to solve for :

$k=m\left ( \frac{2\pi }{T}\right )^{2}=0.3:kg\cdot \left ( \frac{2\pi }{0.25:s} \right )^{2}=189.5:\frac{N}{m}$

Now that we have the force constant, , we can use this to help us find the amplitude of the oscillations. We were also given the energy of the system to be 3.0 J. The equation for the energy of a mass-spring oscillating system is: