Question

A transverse wave is set up in a very long string. The oscillator is set at 20.0 Hz, and the wave speed is 78 m/s. The amplitude of the wave is 5.2 cm. What is the maximum particle acceleration for a point on the string?

Answer 1

To solve this problem we must apply the concepts related to Tangential Acceleration based on angular velocity and acceleration, and therefore, we must also calculate angular velocity based on the given frequency. For all these problems we will take the Units to the International System. The maximum acceleration would then be defined as,

$a_{max} = \omega^2 A$

Here,

$\omega$= Angular velocity

A = Amplitude

At the same time the angular velocity is described as,

$\omega = 2\pi f$

Here f means the frequency of the wave. Substituting,

$\omega = 2 \pi (20)$

$\omega = 40\pi$

$A = 5.2cm$

$A = 0.052m$

Replacing at the first equation,

$a_{max} = (40\pi )^2 (0.052)$

$a_{max} = 821.15m/s^2$

Therefore the maximum particle acceleration for a point on the string is $821.15m/s^2$