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Chapter 14: Problem 29

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

Short Answer

Expert verified
Answer: The maximum amplitude of the oscillation without the upper mass slipping off the lower mass is 0.011772 m.

Step by step solution

01

Determine the maximum friction force between the masses

To find the maximum friction force between the two masses, we can use the formula: \(max\_friction\_force = μ * normal\_force\) where \(μ = 0.600\) (the coefficient of static friction) and \(normal\_force = m_{2}g\), where \(g\) is the gravitational constant, approximately equal to \(9.81 m/s^2\). \(max\_friction\_force = 0.600 * (20.0 * 10^{-3}) * 9.81\)
02

Calculate the maximum friction force between the masses

Now, we can calculate the maximum friction force between the masses: \(max\_friction\_force = 0.600 * 0.020 * 9.81 N = 0.11772 N\)
03

Determine the force exerted by the spring at a certain amplitude

We know that the force exerted by the spring (\(F_{spring}\)) is given by Hooke's law: \(F_{spring} = k * x\) where \(k = 10.0 N/m\) (spring constant) and \(x\) is the displacement (or amplitude), which we are trying to find. For the upper mass not to slip off, \(F_{spring} = max\_friction\_force\)
04

Solve for the maximum amplitude

Now we can set the force exerted by the spring equal to the maximum friction force and solve for the maximum amplitude, \(x\): \(10.0 * x = 0.11772\) \(x = \frac{0.11772}{10.0}\)
05

Calculate the maximum amplitude

Finally, we can calculate the maximum amplitude: \(x = \frac{0.11772}{10.0} = 0.011772 m\) The maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1}\) is 0.011772 m.

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Most popular questions from this chapter

Object \(A\) is four times heavier than object B. Fach object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(A\) is half that of \(B ?\) a) \(T_{A}: T_{\mathrm{g}}=1: 4\) c) \(T_{A}: T_{\mathrm{B}}=2\) : b) \(T_{A}: T_{B}=4: 1\) d) \(T_{A}: T_{B}=1: 2\)

Identical blocks oscillate on the end of a vertical spring one on Earth and one on the Moon. Where is the period of the oscillations greater? a) on Earth d) cannot be determined b) on the Moon from the information given c) same on both Farth and Moon

{~A} 100 \cdot \mathrm{g}\( block hangs from a spring with \)k=5.00 \mathrm{~N} / \mathrm{m}\( At \)t=0 \mathrm{~s},\( the block is \)20.0 \mathrm{~cm}\( below the equilibrium posi. tion and moving upward with a speed of \)200, \mathrm{~cm} / \mathrm{s}\(. What is the block's speed when the displacement from equilibrium is \)30.0 \mathrm{~cm} ?$

A 3.0 -kg mass is vibrating on a spring. It has a resonant angular speed of \(2.4 \mathrm{rad} / \mathrm{s}\) and a damping angular speed of \(0.14 \mathrm{rad} / \mathrm{s}\). If the driving force is \(2.0 \mathrm{~N}\), find the maximum amplitude if the driving angular speed is (a) \(1.2 \mathrm{rad} / \mathrm{s},\) (b) \(2.4 \mathrm{rad} / \mathrm{s}\), and \((\) c) \(4.8 \mathrm{rad} / \mathrm{s}\).

A Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?
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