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Chapter 2: Problem 22

A pendulum whose period in a vacuum is \(1 \mathrm{~s}\) is placed in a resistive medium. Its amplitude on each swing is observed to be half that on the previous swing. What is its new period? Given that the pendulum bob is of mass \(0.1 \mathrm{~kg}\), and is subjected to a periodic force of amplitude \(0.02 \mathrm{~N}\) and period \(1 \mathrm{~s}\), find the angular amplitude of the resulting forced oscillation. Compare your answer with the angular deviation that would be induced by a constant force of this magnitude in a pendulum at rest.

Short Answer

Expert verified
Answer: To compare the angular amplitude of the forced oscillation (\(\theta_m\)) to the angular deviation for a constant force (\(\theta_c\)), you can calculate the percentage difference between these values. The percentage difference is given by: $$\% \: Difference = \frac{|\theta_m - \theta_c|}{\frac{\theta_m + \theta_c}{2}} * 100$$ Substituting the calculated values of \(\theta_m\) and \(\theta_c\) from the previous steps, you will get the percentage difference between the angular amplitude of the forced oscillation and the angular deviation for a constant force.

Step by step solution

01

Calculate the damping ratio

First, let's calculate the damping ratio, "b", using the formula: $$b = \frac{ln(\frac{A_1}{A_2})}{T}$$ Here, \(A_1\) is the amplitude of the first swing, \(A_2\) is the amplitude of the second swing, and \(T\) is the period. Since we know that the amplitude halves on each swing, we can write: $$b = \frac{ln(\frac{A_1}{A_1 / 2})}{1s} = \frac{ln(2)}{1s}$$
02

Calculate the new period

To find the new period of the pendulum, we can use the equation: $$T_d = \frac{T_0}{\sqrt{1 - b^2}}$$ Where \(T_d\) is the damped period, \(T_0\) is the period in a vacuum, and \(b\) is the damping ratio. Plugging in values, we get: $$T_d = \frac{1}{\sqrt{1 - (\frac{ln(2)}{1s})^2}}$$
03

Calculate the angular amplitude of forced oscillation

We can use the following formula to find the angular amplitude (\(\theta_m\)) of the forced oscillation: $$\theta_m = \frac{F_0}{m * g * (1 - \frac{T^2}{T_d^2})^2 + (2 * b * \frac{T}{T_d})^2)}$$ Where \(F_0\) is the amplitude of the periodic force (0.02 N), \(m\) is the mass of the pendulum bob (0.1 kg), \(g\) is the acceleration due to gravity (9.81 m/s²), \(T\) is the period of the periodic force (1s), \(T_d\) is the damped period, and \(b\) is the damping ratio. Substituting values, we get: $$\theta_m = \frac{0.02}{0.1 * 9.81 * (1 - \frac{1^2}{T_d^2})^2 + (2 * \frac{ln(2)}{1s} * \frac{1}{T_d})^2}$$
04

Calculate the angular deviation for a constant force

For a constant force, the angular deviation (\(\theta_c\)) is simply given by: $$\theta_c = \frac{F_c}{m * g}$$ Where \(F_c\) is the constant force (0.02 N), and \(m\) is the mass of the pendulum bob (0.1 kg). Plugging in values, we get: $$\theta_c = \frac{0.02}{0.1 * 9.81}$$
05

Compare angular amplitudes

Now, compare the angular amplitude of the forced oscillation (\(\theta_m\)) with the angular deviation for a constant force (\(\theta_c\)). You can compute the percentage difference between these values to get an idea of how close they are.

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Most popular questions from this chapter

An oscillator with free period \(\tau\) is critically damped and subjected to a force with the 'saw-tooth' form $$ F(t)=c(t-n \tau) \quad \text { for } \quad\left(n-\frac{1}{2}\right) \tau

A particle of mass \(m\) moves (in the region \(x>0\) ) under a force \(F=\) \(-k x+c / x\), where \(k\) and \(c\) are positive constants. Find the corresponding potential energy function. Determine the position of equilibrium, and the frequency of small oscillations about it.

For the pendulum described in \(\S 2.1\), find the equation of motion for small displacements from the position of unstable equilibrium, \(\theta=\pi\). Show that if it is released from rest at a small angle to the upward vertical, then the time taken for the angular displacement to increase by a factor of 10 will be approximately \(\sqrt{l / g} \ln 20\). Evaluate this time for a pendulum of period \(2 \mathrm{~s}\), and find the angular velocity of the pendulum when it reaches the downward vertical.

A particle of mass \(m\) moves under a conservative force with potential energy function given by $$ V(x)= \begin{cases}\frac{1}{2} k\left(a^{2}-x^{2}\right) & \text { for }|x|0\). What is the force on the particle? Sketch the function \(V\), for both cases \(k>0\) and \(k<0\), and describe the possible types of motion.

A particle of mass \(m\) has the potential energy function \(V(x)=m k|x|\) where \(k\) is a positive constant. What is the force when \(x>0\), and when \(x<0 ?\) Sketch the function \(V\) and describe the motion. If the particle starts from rest at \(x=-a\), find the time it takes to reach \(x=a\).
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