Question

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0 s. It then oscillates with a period of 2.00 s and a maximum speed of 60.0 cm/s . You may want to review (Pages 391 - 393) . Part A What is the amplitude of the oscillation? Express your answer with the appropriate units.

Answer 1

Answer:

0.19 m (19 cm)

Explanation:

The maximum speed in a simple harmonic motion is given by

$v = A \omega$ (1)

where

A is the amplitude

$\omega$ is the angular frequency, which is given by

$\omega = \frac{2\pi}{T}$ (2)

where T is the period.

By combining (1) and (2), we find

$v=\frac{2\pi A}{T}\\A=\frac{vT}{2\pi}$

Here we know that

v = 60.0 cm/s = 0.6 m/s is the maximum speed

T = 2.00 s is the period

Substituting into the formula, we find the amplitude:

$A=\frac{(0.6 m/s)(2.0 s)}{2\pi}=0.19 m$