Question 1

How much mass should be attached to a vertical ideal spring having a spring constant of 39.5 N/m so that it will oscillate at 1.00 Hz?

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Question 2

A 1.5-kg cart attached to an ideal spring with a spring constant of 20 N/m oscillates on a horizontal,frictionless track.At time t = 0.00 s,the cart is released from rest at position x = 10 cm from the equilibrium position.
(a)What is the frequency of the oscillations of the cart?
(b)Determine the maximum speed of the cart.Where does the maximum speed occur?
(c)Find the maximum acceleration of the mass.Where does the maximum acceleration occur?
(d)How much total energy does this oscillating system contain?
(e)Express the displacement as a function of time using a cosine function.

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Question 3

An object of mass m = 8.0 kg is attached to an ideal spring and allowed to hang in the earth's gravitational field.The spring stretches $2.2 \mathrm {~cm}$ before it reaches its equilibrium position.If it were now allowed to oscillate by this spring,what would be its frequency?

Accepted Answer

The correct answer is A because the frequency of an object attached to an ideal spring is given by f = (1/2π)√(k/m), where k is the spring constant and m is the mass of the object. In this case, we can find the spring constant by using the equation F = kx, where F is the force applied to the spring, x is the displacement of the spring, and k is the spring constant. The force applied to the spring is equal to the weight of the object, which is mg, where g is the acceleration due to gravity. The displacement of the spring is 2.2 cm, or 0.022 m. Substituting these values into the equation, we get k = mg/x = (8.0 kg)(9.8 m/s^2)/(0.022 m) = 352.7 N/m. Substituting this value of k into the equation for frequency, we get f = (1/2π)√(k/m) = (1/2π)√(352.7 N/m / 8.0 kg) = 3.4 Hz.

Question 4

An object of mass 6.8 kg is attached to an ideal spring of force constant (spring constant)1720 N/m.The object is set into simple harmonic motion,with an initial velocity of $$v _ { \mathrm { O } } = 4.1 \mathrm {~m} / \mathrm { s }$$ and an initial displacement of $x _ { 0 } = 0.11 \mathrm {~m}$ Calculate the maximum speed the object reaches during its motion.

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Question 5

A 0.150-kg cart that is attached to an ideal spring with a spring constant of 3.58 N/m undergoes simple harmonic oscillations with an amplitude of 7.50 cm.What is the total mechanical energy of the system?

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Question 6

A block attached to an ideal spring of spring constant 15 N/m executes simple harmonic motion on a frictionless horizontal surface.At time t = 0 s,the block has a displacement of -0.90 m, a velocity of -0.80 m/s, and an acceleration of +2.9 m/s² .The mass of the block is closest to

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Question 7

A 0.16-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is 360 N/m.The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.When the position is x = -0.037 m,what is the acceleration of the block?

Accepted Answer

The acceleration $a$ of the block in simple harmonic motion can be calculated using Hooke's Law, $F = -kx$, and Newton's second law, $F = ma$, where $k$ is the spring constant, $x$ is the displacement from the equilibrium position, $m$ is the mass of the block, and $a$ is the acceleration. Thus, $a = -\frac{k}{m}x$. Substituting $k = 360 \, \text{N/m}$, $m = 0.16 \, \text{kg}$, and $x = -0.037 \, \text{m}$, we get $a = -\frac{360}{0.16}(-0.037) = 83 \, \text{m/s}^2$.

Question 8

A 0.50-kg object is attached to an ideal spring of spring constant 20 N/m along a horizontal,frictionless surface.The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.At what distance from the equilibrium position are the kinetic energy and potential energy of the system the same?

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Question 9

A geologist suspends a 0.30-kg stone on an ideal spring.In equilibrium the stone stretches the spring 2.0 cm downward.The stone is then pulled an additional distance of 1.0 cm down and released from rest.
(a)Write down the equation for the vertical position y of the stone as a function of time t,using the cosine function.Take the origin at the equilibrium point of the stone,with the positive y direction upward.
(b)How fast is the stone moving at a time equal to 1/3 of its period of motion?

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Question 10

A 0.50-kg object is attached to an ideal spring of spring constant 20 N/m along a horizontal,frictionless surface.The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.What are (a)the total energy and (b)the amplitude of vibration of the system?

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Question 11

A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is $$570 \mathrm {~N} / \mathrm { m } .$$ The block is pulled from its equilibrium position at x = 0.000 m to a displacement x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.When the position of the block is $x = 0.057 \mathrm {~m}$ its kinetic energy is closest to

Accepted Answer

The correct answer is A because the kinetic energy of the block is maximum when it passes through its equilibrium position. The kinetic energy of the block is given by:$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2\sin^2(\omega t)$$where m is the mass of the block, v is the velocity of the block, $\omega$ is the angular frequency of the motion, A is the amplitude of the motion, and t is the time.At x = 0.057 m, the block is at a displacement of A/2 from its equilibrium position. Therefore, the kinetic energy of the block is:$$K = \frac{1}{2}m\omega^2A^2\sin^2(\omega t) = \frac{1}{2}(0.39 \mathrm{~kg})(570 \mathrm{~N} / \mathrm{m})(0.080 \mathrm{~m})^2\sin^2(\omega t) = 0.90 \mathrm{~J}$$

Question 12

What is the length of a simple pendulum with a period of 2.0 s?

Accepted Answer

The formula for the period of a simple pendulum is T = 2π√(L/g), where T is the period in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity (9.8 m/s²). Solving for L, we get L = (gT²)/(4π²) = (9.8 m/s²)(2.0 s)²/(4π²) ≈ 0.99 m. Therefore, the best choice is B.

Question 13

A 2.0 kg box is traveling at 5.0 m/s on a smooth horizontal surface when it collides with and sticks to a stationary 6.0 kg box.The larger box is attached to an ideal spring of force constant (spring constant)150 N/m,as shown in the figure.Find (a)the amplitude of the resulting oscillations of this system,(b)the frequency of the oscillations and (c)the period of the oscillations.

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Question 14

A 0.50-kg box is attached to an ideal spring of spring constant 20 N/m on a horizontal,frictionless floor.The box oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
(a)What is the amplitude of vibration?
(b)At what distance from the equilibrium position are the kinetic energy and the potential energy the same?

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Question 15

A 0.30-kg block of wood is suspended on a spring.In equilibrium the wood stretches the spring 2.0 cm downward.The wood is then pulled an additional distance of 1.0 cm down and released from rest.
(a)How long does it take the wood to make 3 complete cycles of vibration?
(b)How much total mechanical energy does this system contain if we choose the total potential energy (elastic and gravitational)to be zero at the equilibrium position of the hanging block?

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Question 16

A 34-kg child on an 18-kg swing set swings back and forth through small angles.If the length of the very light supporting cables for the swing is $4.9 \mathrm {~m}$ how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables.

Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity ($9.8 \mathrm{~m/s^2}$). Plugging in the given length of $4.9 \mathrm{~m}$, we get $T = 2\pi\sqrt{\frac{4.9}{9.8}} = 2\pi\sqrt{0.5} = 2\pi\sqrt{\frac{1}{2}} = 2\pi(\frac{1}{\sqrt{2}}) \approx 4.4 \mathrm{~s}$. The mass of the child and swing set does not affect the period of the swing for small angles, so the answer is 4.4 seconds.

Question 17

A ball is attached to an ideal spring and oscillates with a period T.If the mass of the ball is doubled,what is the new period?

Accepted Answer

The period of a mass-spring system is given by $ T = 2\pi \sqrt{\frac{m}{k}} $. If the mass $ m $ is doubled, the new period $ T' $ becomes $ 2\pi \sqrt{\frac{2m}{k}} = T \sqrt{2} $.

Question 18

A ball vibrates back and forth from the free end of an ideal spring having a spring constant of 20 N/m.If the amplitude of this motion is 0.30 m,what is the kinetic energy of the ball when it is 0.30 m from its equilibrium position?

Accepted Answer

When the ball is 0.30 m from its equilibrium position, it is at its maximum displacement (amplitude), where all the energy is potential and the kinetic energy is zero.

Question 19

A 3.7-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is $450 \mathrm {~N} / \mathrm { m }$ The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.The maximum elastic potential energy of the system is closest to

Accepted Answer

The maximum elastic potential energy of the system is equal to the work done in stretching the spring from its equilibrium position to its maximum displacement. The work done is given by:$$W = \frac{1}{2}kx^2$$where k is the spring constant and x is the displacement from equilibrium. Substituting the given values, we get:$$W = \frac{1}{2}(450 \mathrm {~N} / \mathrm { m })(0.080 \mathrm {~m})^2 = 1.44 \mathrm {~J}$$Therefore, the maximum elastic potential energy of the system is closest to 1.4 J.

Question 20

A 1.53-kg piece of iron is hung by a vertical ideal spring.When perturbed slightly,the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm.If we choose the total potential energy (elastic and gravitational)to be zero at the equilibrium position of the hanging iron,what is the total mechanical energy of the system?

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Quiz 20: Vibrational Motion