Period and Frequency of Harmonic Motion - AP Physics 1

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Question

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?

Answer

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:

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

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Question

Consider the following system:

If the period of the pendulum is , what is the length of the string?

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

Answer

We simply need the expression for the period of a pendulum to solve this problem:

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

Rearranging for length, we get:

$l = g\left ( \frac{T}{2\pi}\right )^2$

We have all of these values, allowing us to solve:
$l = (10\frac{m}{s^2})\left ( \frac{5s}{2\pi}\right )^2 = 6.3m$

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Question

Consider the following system:

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

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

Answer

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:

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

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Question

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?

Answer

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.

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

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Question

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?

Answer

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:

$PE_{s}=\frac{1}{2}kx^{2}$

is the total potential energy of the system. Now, plug in values and solve for , which is the amplitude of the oscillations.

$x=\sqrt{\frac{2\cdot PE_s}{k}}$

$x=\sqrt{\frac{2\cdot 3.0:J}{189.5:\frac{N}{m}}}=0.18:m$

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Question

A pendulum of length undergoing simple harmonic motion has a period of 1.0 s on Earth. If the acceleration of gravity on the Moon is $1.6 :\frac{m}{s^2}$ , what would the period of the pendulum be on the Moon?

$g_{earth}=9.8 :\frac{m}{s^2}$

Answer

The equation that relates the period of a pendulum and its length is

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

We know the period of the pendulum on Earth, which was given, to be 1.0 s. We also know that on Earth is $9.8 :\frac{m}{s^2}$ . So, we can find the length of the pendulum:

$l=g_{earth}\left (\frac{T_{earth}}{2\pi } \right )^{2}=9.8:\frac{m}{s^2}\left ( \frac{1.0:s}{2\pi } \right )^{2}=0.248:m$

Now that we have the length of the pendulum, we can use this to calculate what the period of oscillation will be if the pendulum was on the Moon where the "g" is not the same as on Earth, but rather is $1.6 :\frac{m}{s^2}$ .

$T_{moon}=2\pi\sqrt{\frac{l}{g_{moon}}}=2\pi\sqrt{\frac{0.248:m}{1.6:\frac{m}{s^2}}}=2.5:s$

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Question

If a period of a pendulum is , what is its length?

Answer

Equation for period of a pendulum:

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

Solve.

$7s = 2\pi \sqrt{\frac{L}{10\frac{m}{s^2}}}$

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Question

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 .

Answer

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.

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Question

In the lab, a student has created an oscillator by hanging a weight from a spring. The student releases the oscillator from rest and uses a sensor and computer to find the equation of motion for the oscillator: $x(t)=0.13m;cos(6.3\frac{radians}{s}t+0.14:radians)$

The student then replaces the weight with a weight whose mass, is twice as large as that of the original weight.The student again releases the weight from rest from the same displacement from equilibrium. What would the new equation of motion be for the oscillator?

Answer

The equation of motion for an oscillator is:

$x(t)=Acos(\omega t+\phi)$

In words, the position as a function of time equals the amplitude times the cosine of the sum of the frequency (in radians per second) times time plus phase. Changing the mass of the weight only changes the frequency, but it depends upon the square root of the mass. Since increasing the mass increases the period, it decreases the frequency by the factor of $\sqrt{2}=1.41$ .

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Question

The Fourier Transform is an extensively used mathematical analysis technique. In some applications, it reconstructs a function into an infinite series of sine waves.

Given that the Fourier Transform is given by the series:

$\sum_{n=1}^{\infty } c_n*sin(\frac{n \pi x}{l})$

Where is an arbitrary length and is an integer, what is the wavelength of the sinusoid when ?

Answer

First, we want to ignore the summation sign and the since those terms do not affect the wavelength at . All we need to look at is the sine term.

For , we get the sinusoid:

$sin(\frac{3 \pi x}{l})$

Remember that wavelength $\lambda$ is given by

$\lambda = \frac{2 \pi}{\frac{3\pi}{l}}= \frac{2l}{3}$

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Question

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

Answer

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

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

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Question

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.

Answer

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

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Question

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?

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

Answer

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:

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

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Question

If the length of a simple pendulum is halved and the pendulum is moved to the moon where $g_{moon}=1.6\ \frac{\textup{m}}{\textup{s}^2}$ , by what factor does the period of the pendulum change when this is done?

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

Answer

Since we are told that this is a simple pendulum, we can use the follow expression for period:

$T_1=2\pi \sqrt{\frac{L_1}{g_1}}$

$T_2=2\pi \sqrt{\frac{L_2}{g_2}}$

Now let's divide scenario 2 by scenario 1:

$\frac{T_2}{T_1}= \sqrt{\frac{L_2g_1}{L_1g_2}}$

From the problem statement, we know that:

So let's plug that in:

$\frac{T_2}{T_1}= \sqrt{\frac{2L_1g_1}{L_1g_2}} = \sqrt{\frac{2g_1}{g_2}}$

Then plugging in our values for g:

$\frac{T_2}{T_1} = \sqrt{\frac{2\left ( 10\frac{m}{s^2}\right )}{\left ( 1.6\frac{m}{s^2}\right )}}$

$\frac{T_2}{T_1} = 3.5$

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Question

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

Answer

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.

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Question

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 frequency of the system?

Answer

In these types of problems, it is always advantageous to recognize the format of the equation. In trigonometric functions, the period is always given by $\frac{2\pi}{B}$ , when the function is written as . Since frequency is the reciprocal of the period, we will need to flip the fraction.

$B=4\pi$

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

$Frequency=\frac{1}{P}=\frac{1}{\frac{1}{2}}=2Hz$

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Question

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?

Answer

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!

$B=4\pi$

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

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Question

A horizontal spring is oscillating with a mass sliding on a perfectly frictionless surface. If the amplitude of the oscillation is and the mass has a value of and a velocity at the rest length of $2\frac{m}{s}$ , determine the frequency of oscillation.

Answer

Using conservation of energy:

Plugging in values:

Solving for

$k=\frac{80N}{m}$

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

Plugging in values:

$\omega=\frac{40 rad}{s}$

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