Question

A mass-and-spring system oscillates with amplitude \(A\) and angular frequency \(\omega\) (a) What is the average speed during one complete cycle of oscillation? (b) What is the maximum speed? (c) Find the ratio of the average speed to the maximum speed. (d) Sketch a graph of \(v_{x}(t),\) and refer to it to explain why this ratio is greater than \(\frac{1}{2}\).

Short answer

In this problem, we have a mass-and-spring system oscillating with amplitude A and angular frequency ω. The velocity formula for the system is given by \(v_x(t) = -A\omega\sin(\omega t + \phi)\). The maximum speed is obtained when the sine function has its maximum value, which is 1. Thus, the maximum speed is given by \(v_\text{max} = A\omega\). The average speed of one complete cycle is the total distance covered over the total time taken, which can be calculated as \(v_\text{avg} = \frac{4A\omega}{2\pi}\). The ratio of average speed to maximum speed is \(\frac{v_\text{avg}}{v_\text{max}} = \frac{2}{\pi}\). The graph of \(v_x(t)\) represents a sine wave with a maximum value of \(A\omega\) and a minimum value of \(-A\omega\), oscillating about the t-axis. The average speed is greater than half of the maximum speed because it takes into account the symmetric slopes of the sine wave in both the positive and negative half cycles.

Step-by-step solution

Displacement formula for oscillatory motion

For a mass-and-spring system undergoing simple harmonic motion, the displacement \(x(t)\) is given by the formula: \(x(t) = A\cos(\omega t + \phi)\) where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle.

Calculate the velocity formula

To find the velocity formula, we need to differentiate the displacement formula with respect to time. So, \(v_x(t) = \frac{dx(t)}{dt} = -A\omega\sin(\omega t + \phi)\)

Find the maximum speed

The maximum speed occurs when the sine function has its max value, which is 1. Hence, the maximum speed of the mass-and-spring system is: \(v_\text{max} = A\omega\)

Find the average speed of one complete cycle

When the system completes one full oscillation, it covers a total distance of \(4A\). The time taken for one full oscillation is the period, given by \(T= \frac{2\pi}{\omega}\). The average speed is the total distance covered divided by the total time taken, so: \(v_\text{avg} = \frac{4A}{T} = \frac{4A\omega}{2\pi}\)

Calculate the ratio of average speed to maximum speed

The ratio of the average speed to the maximum speed is given by: \(\frac{v_\text{avg}}{v_\text{max}} = \frac{4A\omega/2\pi}{A\omega} = \frac{2}{\pi}\)

Sketch the graph of \(v_x(t)\) and explain the ratio

The graph of \(v_x(t)\) is a sine wave with a maximum value of \(A\omega\) and minimum value of \(-A\omega\), oscillating about the t-axis and crossing it at points \(2\pi k\) for any integer \(k\). The graph shows that the average speed is greater than half of the maximum speed. This is because the average speed takes into account both the positive and negative half cycles, and the positive and negative slopes of the sine wave are symmetric about the t-axis. Therefore, the average speed is greater than \(\frac{1}{2}\) of the maximum speed.