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

0.19 m (19 cm)

Step-by-step 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 = (2\pi)/(T)` (2)

where T is the period.

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

`v=(2\pi A)/(T)\\A=(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=((0.6 m/s)(2.0 s))/(2\pi)=0.19 m`