Question 1

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be located?

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Question 2

An object of mass of 2.0 kg hangs from an ideal massless spring with a spring constant of 50 N/m. An oscillating force F = (4.8 N) cos[(3.0 rad/s)t] is applied to the object. What is the amplitude of the resulting oscillations? You can neglect damping.

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Question 3

A 5.0-kg block is attached to an ideal massless spring whose spring constant is 125 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a position at x = +0.687 m and is released from rest. The block then executes lightly damped oscillation along the x-axis, and the damping force is proportional to the velocity. When the block first returns to x = 0.00 m, its x component of velocity is -2.0 m/s and its x component of acceleration is +5.6 m/s². (a) What is the magnitude of the acceleration of the block upon release at x = +0.687 m. (b) Find the damping constant b.

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Question 4

A uniform meter stick is freely pivoted about the 0.20-m mark. If it is allowed to swing in a vertical plane with a small amplitude and friction, what is the frequency of its oscillations?

Accepted Answer

The frequency of oscillation of a physical pendulum like a meter stick pivoted at a point other than its center of mass can be found using the formula $f = \frac{1}{2\pi}\sqrt{\frac{g}{l_{\text{eff}}}}$, where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$) and $l_{\text{eff}}$ is the effective length from the pivot point to the center of mass of the pendulum. For a uniform meter stick pivoted at the 0.20 m mark, the center of mass is at the 0.50 m mark from one end, making it 0.30 m from the pivot. Plugging these values into the formula gives a frequency close to 0.66 Hz.

Question 5

A 2.15 kg lightly damped harmonic oscillator has an angular oscillation frequency of 0.261 rad/s. If the maximum displacement of 2.0 m occurs when t = 0.00 s, and the damping constant b is 0.74 kg/s what is the object's displacement when t = 4.01 s?

Accepted Answer

The correct answer is A because the displacement of the object at time t = 4.01 s is 0.50 m. This can be calculated using the equation:x(t) = A * e^(-bt/2m) * cos(ωt + φ)where:x(t) is the displacement of the object at time tA is the maximum displacement of the objectb is the damping constantm is the mass of the objectω is the angular oscillation frequencyφ is the phase anglePlugging in the given values, we get:x(4.01 s) = 2.0 m * e^(-0.74 kg/s * 4.01 s / 2 * 2.15 kg) * cos(0.261 rad/s * 4.01 s + 0)= 0.50 m

Question 6

An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b?

Accepted Answer

The damping constant b can be calculated using the formula:$$b = \frac{m}{Q}\frac{\Delta A}{\Delta t}$$where m is the mass of the object, Q is the quality factor, ΔA is the change in amplitude, and Δt is the time interval.The quality factor Q is given by:$$Q = \frac{2\pi f_0}{\Delta f}$$where f_0 is the resonant frequency and Δf is the bandwidth.The resonant frequency is given by:$$f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$where k is the spring constant.The bandwidth is given by:$$\Delta f = \frac{b}{2\pi m}$$Substituting these equations into the formula for b, we get:$$b = \frac{m}{Q}\frac{\Delta A}{\Delta t} = \frac{m}{2\pi f_0}\frac{\Delta A}{\Delta t}\frac{b}{2\pi m} = \frac{\Delta A}{2\pi \Delta t}\frac{m}{f_0}$$Substituting the given values, we get:$$b = \frac{5.00 	imes 10^{-3} 	ext{ m}}{2\pi 	imes 15.0 	ext{ s}}\frac{0.0750 	ext{ kg}}{1.26 	ext{ Hz}} = 0.00693 	ext{ kg/s}$$Therefore, the magnitude of the damping constant b is 0.00693 kg/s.

Question 7

A 25 kg object is undergoing lightly damped harmonic oscillations. If the maximum displacement of the object from its equilibrium point drops to 1/3 its original value in 1.8 s, what is the value of the damping constant b?

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Question 8

A lightly damped harmonic oscillator, with a damping force proportional to its speed, is oscillating with an amplitude of 0.500 cm at time t = 0. When t = 8.20 s, the amplitude has died down to 0.400 cm. At what value of t will the oscillations have an amplitude of 0.250 cm?

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Question 9

A large stick is pivoted about one end and allowed to swing back and forth with no friction as a physical pendulum. The mass of the stick is 5.40 kg and its center of gravity (found by finding its balance point) is 1.80 m from the pivot. If the period of the swinging stick is 6.90 seconds, what is the moment of inertia of the stick about an axis through the pivot?

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Question 10

The amplitude of a lightly damped harmonic oscillator decreases from 60.0 cm to 40.0 cm in 10.0 s. What will be the amplitude of the harmonic oscillator after another 10.0 s passes?

Accepted Answer

The amplitude of a lightly damped harmonic oscillator decreases exponentially over time. Given that the amplitude decreases from 60.0 cm to 40.0 cm in 10.0 s, it means that after another 10.0 s, the amplitude will decrease by the same factor, which is 40.0/60.0 = 2/3. Therefore, after another 10.0 s, the amplitude will be 2/3 of 40.0 cm, which is approximately 26.7 cm.

Quiz 13: Periodic Motion