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AP Physics 1 : Other Harmonic Systems

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Example Questions

Example Question #1 : Other Harmonic Systems

You push on a door with a force of 1.3 N at an angle of 45 degrees to the surface and at a distance of 0.5 m from the hinges. What is the torque produced?

Possible Answers:

$0.46N\cdot m$

$0.64N\cdot m$

$0.31N\cdot m$

$0.84N\cdot m$

$1.2N\cdot m$

Correct answer:

$0.46N\cdot m$

Explanation:

To calculate torque, this equation is needed:

$\tau=rFsin\Theta$

Next, identify the given information:

r= 0.5 m

F=1.3N

$\Theta=45^{o}$

Plug these numbers into the equation to determine the torque:

$\tau=rFsin\Theta = (0.5m)(1.5N)sin(45^{o})=0.46N\cdot m$

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Example Question #43 : Circular, Rotational, And Harmonic Motion

Photo_17

If a mass on the end of a string of length 8 cm is pulled 30 degrees away from vertical, what will its speed be when the string is vertically alligned if the mass is released from rest?

$g=9.81\frac{m}{s^2}$

Possible Answers:

$2.1\frac{m}{s}$

$1.3\frac{m}{s}$

$4.5\frac{m}{s}$

$1.7\frac{m}{s}$

$0.46\frac{m}{s}$

Correct answer:

$0.46\frac{m}{s}$

Explanation:

We can determine the vertical components by using some geometry.

Photo_2

Since this problem boils down to being a problem of conservation of energy, we can state that

$E_{i}=E_{f}\rightarrow PE_{i}+KE_{i}=PE_{f}+KE_{f}$

Since the mass is released from rest and we can state at the bottom of its arc it will be at height 0 m, this equation can be simplified:

$PE_{i}=KE_{f} \rightarrow mgh=\frac{1}{2}mv^{2} \rightarrow gh=\frac{1}{2}v^{2}$

is equal to the height of the mass above 0 m; therefore, by referring to the diagram, we can determine that $h=R-Rcos\Theta$ . By plugging this in and rearranging the equation, we can solve for the speed of the mass when the string is vertically aligned:

$v=\sqrt{2gR(1-cos\Theta)}=\sqrt{2(9.81\frac{m}{s^2})(8\cdot10^{-2}m)(1-cos30^{o})}=0.46\frac{m}{s}$

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Example Question #41 : Harmonic Motion

A ball with mass 5 kg is attached to a spring and is released 10 meters from equilibrium. After some time, the ball passes the equilibrium point moving at $7\frac{m}{s}$ . What is the spring constant () of the spring?

Possible Answers:

$2.45 \frac{kg}{s^2}$

$1.225\frac{kg}{s^2}$

$0.41\frac{kg}{s^2}$

$3.5\frac{kg}{s^2}$

$0.82\frac{kg}{s^2}$

Correct answer:

$2.45 \frac{kg}{s^2}$

Explanation:

When the object is released, it has no kinetic energy (it isn't moving) and a potential energy of

$P.E.=\frac{1}{2}kx^2=\frac{1}{2}k(10m)^2=50m^2\cdot k$

When the object passes through the equilibrium point, it has no potential energy ( x=0 ) and a kinetic energy of

$K.E.=\frac{1}{2}mv^2=\frac{1}{2}(5kg)(7\frac{m}{s})^2=122.5\frac{kg\cdot m^2}{s^2}$

Due to the conservation of energy, these two quantities must be equal to each other:

$50m^2\cdot k=122.5\frac{kg*m^2}{s^2}$

$k = \frac{122.5\frac{kg\cdot m^2}{s^2}}{50m^2}=2.45\frac{kg}{s^2}$

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Example Question #1 : Other Harmonic Systems

A 30 kg weight attached to a spring is at equilibrium lying horizontally on a table. The spring is lifted up and is stretched by 80 cm before the weight is lifted off the table. What is this spring's spring constant ()?

Possible Answers:

$0.375\frac{kg}{s^2}$

$460\frac{kg}{s^2}$

$37.5\frac{kg}{s^2}$

$3.68\frac{kg}{s^2}$

$368\frac{kg}{s^2}$

Correct answer:

$368\frac{kg}{s^2}$

Explanation:

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

F_g=mg

$294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

-kx=-294N

k(0.8m) = 294N

$k=368\frac{N}{m}=368\frac{kg}{s^2}$

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Example Question #1 : Other Harmonic Systems

Instead of using a second hand to count seconds, a watchmaker decides to construct a simple harmonic system involving a mass and a spring suspended from a table.

Ignoring the effects of gravity, if the spring the watchmaker selects has a spring constant of $0.01\frac{kg}{s^2}$ , how large of a mass in kilograms should he attach to the spring such that the harmonic system does not oscillate too slow or too fast?

Possible Answers:

$\frac{1}{3}kg$

$\frac{3}{\pi^2}kg$

$\frac{9}{10}kg$

$\frac{9}{\pi^2}kg$

$\frac{1}{6}kg$

Correct answer:

$\frac{9}{\pi^2}kg$

Explanation:

The angular velocity of the harmonic system is equal to the square root of the spring constant over the mass:

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

Since we need the angular velocity to be $\frac{2\pi}{60}$ and we are given the spring constant as $0.01\frac{kg}{s^2}$ , then we can set up an equation:

$\frac{2\pi}{60}=\sqrt{\frac{0.01}{m}}$

$\frac{4\pi^2}{60^2}=\frac{0.01}{m}$

$m=\frac{60^2*0.01}{4\pi^2}=\frac{9}{\pi^2}kg$

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Example Question #5 : Other Harmonic Systems

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 velocity equation of the system as a function of time?

Possible Answers:

$v(t)=8\pi cos( t)$

$v(t)=2 cos(4\pi t)$

$v(t)=2 sin(4\pi t)$

$v(t)=8\pi sin(4\pi t)$

$v(t)=8\pi cos(4\pi t)$

Correct answer:

$v(t)=8\pi cos(4\pi t)$

Explanation:

The velocity equation can be found by differentiating the position equation.

$v(t)=x'(t)=\frac{d}{dt}(2sin(4\pi t)+4)=2*4\pi cos(4\pi t)$

$=8\pi cos(4\pi t)$

Here, we made use of the chain rule in taking the derivative.

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AP Physics 1 : Other Harmonic Systems

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Other Harmonic Systems

Example Question #43 : Circular, Rotational, And Harmonic Motion

Example Question #41 : Harmonic Motion

Example Question #1 : Other Harmonic Systems

Example Question #1 : Other Harmonic Systems

Example Question #5 : Other Harmonic Systems

All AP Physics 1 Resources

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