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AP Physics 1 : Harmonic Motion

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Example Question #11 : Period And Frequency Of Harmonic Motion

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?

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

Possible Answers:

0.57

0.28

3.5

1.0

1.8

Correct answer:

3.5

Explanation:

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:

L_2 = 2L_1

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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Example Question #61 : Harmonic Motion

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

Possible Answers:

min U, min K

max U, min K

None of these

max U, max K

min U, max K

Correct answer:

min U, max K

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.

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Example Question #62 : Harmonic Motion

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 .

Possible Answers:

5kg

Impossible to determine

7.90kg

$5.07*10^{-3}kg$

3.17kg

Correct answer:

Impossible to determine

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.

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Example Question #1091 : Newtonian Mechanics

The position of a 1kg mass in an oscillating spring-mass system is given by the following equation:

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

What is the frequency of the system?

Possible Answers:

$\frac{1}{2}Hz$

$\frac{1}{4}Hz$

8Hz

4Hz

2 Hz

Correct answer:

2 Hz

Explanation:

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 Asin(Bt+C)+D . 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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Example Question #191 : Circular, Rotational, And Harmonic Motion

The position of a 1kg mass in an oscillating spring-mass system is given by the following equation:

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

What is the period of the oscillations?

Possible Answers:

$\frac{1}{2}s$

$\frac{1}{4}s$

Correct answer:

$\frac{1}{2}s$

Explanation:

In trigonometric functions, the period is always given by $\frac{2\pi}{B}$ , when the function is written as Asin(Bt+C)+D . 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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Example Question #61 : Harmonic Motion

A horizontal spring is oscillating with a mass sliding on a perfectly frictionless surface. If the amplitude of the oscillation is 5cm and the mass has a value of 50g and a velocity at the rest length of $2\frac{m}{s}$ , determine the frequency of oscillation.

Possible Answers:

$\frac{80 rad}{s}$

None of these

$\frac{395 rad}{s}$

$\frac{710 rad}{s}$

$\frac{40 rad}{s}$

Correct answer:

$\frac{40 rad}{s}$

Explanation:

Using conservation of energy:

KE_1+EPE_1=KE_2+EPE_2

.5mv_1^2+.5kx_1^2=.5mv_2^2+.5kx^2

Plugging in values:

.5*.050*2^2+.5k(0)^2=.5*.050(0)^2+.5k*.05^2

Solving for

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

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

Plugging in values:

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

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AP Physics 1 : Harmonic Motion

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #11 : Period And Frequency Of Harmonic Motion

Example Question #61 : Harmonic Motion

Example Question #62 : Harmonic Motion

Example Question #1091 : Newtonian Mechanics

Example Question #191 : Circular, Rotational, And Harmonic Motion

Example Question #61 : Harmonic Motion

All AP Physics 1 Resources

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