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High School Physics : Harmonic Motion

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Example Questions

Example Question #21 : Harmonic Motion

A spring with a spring constant of $325\frac{N}{m}$ has a mass of 0.22kg attached to one end. It is stretched a distance of {0.64m} . How much force is required to restore the spring to its equilibrium position?

Possible Answers:

-66.56N

-133.12N

-208N

-104N

-71.5N

Correct answer:

-208N

Explanation:

The formula for the restoring force of a spring is:

$F=-k\Delta x$

Essentially, the restoring force is equal and opposite to the force required to stretch the spring. Note that the mass has no place in this calculation. We are given the spring constant and displacement, allowing us to calculate the force.

$F=-k\Delta x$

$F=-325\frac{N}{m}*0.64m$

F=-208N

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Example Question #12 : Using Spring Equations

A spring with a spring constant of $180\frac{N}{m}$ is compressed 0.22m . What is the potential energy in the spring?

Possible Answers:

0.05J

8.71J

9.21J

4.36J

19.8J

Correct answer:

4.36J

Explanation:

The formula for potential energy in a spring is:

$PE=\frac{1}{2}kx^2$

We are given the values for the spring constant and displacement, allowing us to calculate the potential energy.

$PE=\frac{1}{2}(180\frac{N}{m})(0.22m)^2$

$PE=\frac{1}{2}(180\frac{N}{m})(0.0484m^2)$

PE=4.36J

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Example Question #13 : Using Spring Equations

How much force is required to compress a spring 0.22m if it has a spring constant of $180\frac{N}{m}$ ?

Possible Answers:

4.36N

19.8N

219.6N

818.18N

39.6N

Correct answer:

39.6N

Explanation:

The formula for compression force in a spring is:

F=kx

We are given the value for the spring constant and the displacement, allowing us to solve for the force required.

$F=180\frac{N}{m}*0.22m$

F=39.6N

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Example Question #22 : Harmonic Motion

A spring is stretched 1.32m in the horizontal direction. If the spring requires -800N of force to restore it to its original position, what is the spring constant?

Possible Answers:

$1056\frac{N}{m}$

$606.1\frac{N}{m}$

$800\frac{N}{m}$

$356.4\frac{N}{m}$

$459.1\frac{N}{m}$

Correct answer:

$606.1\frac{N}{m}$

Explanation:

To solve this problem, use Hooke's law.

F=-kx

We know the force of the spring and the distance it is displaced. Using these values, we can solve for the spring constant.

-800N=-(k)(1.32m)

$\frac{-800N}{1.32m}=-k$

$-606.1\frac{N}{m}=-k$

$606.1\frac{N}{m}=k$

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Example Question #23 : Harmonic Motion

A spring with a spring constant of $1500\frac{N}{m}$ is compressed 0.87m . A 0.67kg object is placed at the end of the compressed spring and the spring is released. What is the maximum velocity of the object?

Possible Answers:

There is insufficient information to solve

$1694.6\frac{m}{s}$

$13.79\frac{m}{s}$

$41.17\frac{m}{s}$

$190.79\frac{m}{s}$

Correct answer:

$41.17\frac{m}{s}$

Explanation:

For this problem, use the law of conservation of energy. Assuming no other forces are acting upon the object the initial spring potential energy will be equal to the maximum final kinetic energy.

$PE_{spring}=KE_f$

Expand this equality with the formulas for each type of energy.

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$

We are given the spring constant and displacement, allowing us to complete the left side of the equation. We are also able to plug in the mass to the left side of the equation.

$\frac{1}{2}(1500\frac{N}{m})(0.87m)^2=\frac{1}{2}(0.67kg)v^2$

Solve to isolate and solve for the velocity variable.

$(750\frac{N}{m})(0.7569m^2)=(0.335kg)v^2$

567.68J=(0.335kg)v^2

$\frac{567.68J}{0.335kg}=v^2$

$1694.57\frac{m^2}{s^2}=v^2$

$\sqrt{1694.57\frac{m^2}{s^2}}=\sqrt{v^2}$

$41.17\frac{m}{s}=v$

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Example Question #16 : Using Spring Equations

How much potential energy is generated by a spring with a spring constant of $820\frac{N}{m}$ if it is stretched 0.79m from equilibrium?

Possible Answers:

511.8J

209.8J

295.5J

647.8J

255.9J

Correct answer:

255.9J

Explanation:

Spring potential energy is equal to half of the spring constant times the compression/stretching distance squared:

$PE=\frac{1}{2}k(\Delta x)^2$

Using the given values for the spring constant and displacement, we can solve for the energy.

$PE=\frac{1}{2}(820\frac{N}{m})(0.79m)^2$

$PE=(410\frac{N}{m})(0.6241m^2)$

PE=255.9J

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Example Question #1 : Spring Force

A 36kg girl bounces on a massless pogo stick. If the spring constant for the stick is $15,000\frac{N}{m}$ , what is the maximum compression of the spring?

$g=-9.8\frac{m}{s^2}$

Possible Answers:

0.02m

0.14m

0.0004m

42.52m

0.2m

Correct answer:

0.02m

Explanation:

There are two forces at work here: the force due to gravity and the restoring force of the spring. We can set these two forces equal to one another because the forces must be in equilibrium when the spring is compressed at its maximum point.

F_g=F_s

Expand this equation by using the formulas for gravitational and spring force, respectively.

mg=-kx

Plug in our given values for the girl's mass, gravitational acceleration, and the spring constant. Using these values, we can solve for the displacement of the spring.

$(36kg)(-9.8\frac{m}{s^2})=-(15,000\frac{N}{m})(x)$

$-352.8N=-15,000\frac{N}{m}*x$

$\frac{-352.8N}{-15,000\frac{N}{m}}=x$

0.02m=x

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Example Question #1 : Using Pendulum Equations

A 1kg mass on a pendulum is dropped from rest. It is observed that the period of the oscillation is . What is the length of the string?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

3.33m

8.94m

0.92m

9.99m

0.955m

Correct answer:

8.94m

Explanation:

The equation for the period of a pendulum is $T=2\pi\sqrt{\frac{L}{g}}$ .

The mass on the pendulum is irrelevant when calculating the relationship between period and string length, as the only variables in the equation are the length of the string and gravity.

Since we are given the period and the constant for gravity, we can solve for the string length.

$6s=2\pi\sqrt{\frac{l}{9.8\frac{m}{s^2}}}$

$\frac{6s}{2\pi}=\sqrt{\frac{l}{9.8\frac{m}{s^2}}}$

$0.955s=\sqrt{\frac{l}{9.8\frac{m}{s^2}}}$

$(0.955s)^2=(\sqrt{\frac{l}{9.8\frac{m}{s^2}}})^2$

$0.912s^2=\frac{l}{9.8\frac{m}{s^2}}$

$0.912s^2* 9.8\frac{m}{s^2}=l$

8.94m=l

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Example Question #2 : Using Pendulum Equations

A 5kg mass on a pendulum is dropped from rest. If the length of the string is 20m , what is the period of oscillation?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

4.49s

9.8s

19.8s

8.98s

20s

Correct answer:

8.98s

Explanation:

The equation for the period of a pendulum is $T=2\pi\sqrt{\frac{L}{g}}$ .

The mass on the pendulum is irrelevant when calculating the relationship between period and string length, as the only variables in the equation are the length of the string and gravity.

Since we are given the string length and the constant for gravity, we can solve for the period.

$T=2\pi\sqrt{\frac{20m}{9.8\frac{m}{s^2}}}$

$T=2\pi\sqrt{2.04s^2}$

$T=2\pi( 1.43s)$

T=8.98s

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Example Question #2 : Using Pendulum Equations

A pendulum of length is attached to a mass of 8kg . As it swings back and forth, what is the period?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

5.68s

1.79s

1.14s

3.48s

11.36s

Correct answer:

3.48s

Explanation:

The formula for period of a pendulum is:

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

We are given the length of the string and the acceleration due to gravity. We are also given the mass, but this information is irrelevant; period is independent of mass.

Use the given values to solve for the period.

$T=2\pi\sqrt{\frac{3m}{9.8\frac{m}{s^2}}}$

$T=2\pi \sqrt{0.31s^2}$

$T=2\pi* 0.55s$

T=3.48s

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High School Physics : Harmonic Motion

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #21 : Harmonic Motion

Example Question #12 : Using Spring Equations

Example Question #13 : Using Spring Equations

Example Question #22 : Harmonic Motion

Example Question #23 : Harmonic Motion

Example Question #16 : Using Spring Equations

Example Question #1 : Spring Force

Example Question #1 : Using Pendulum Equations

Example Question #2 : Using Pendulum Equations

Example Question #2 : Using Pendulum Equations

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