Question

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 cm/sec and the period is 628 milli-seconds. The amplitude of the motion in cm is:_______.

a. 3.
b. 2.
c. 1.5.
d. 1.0.

Answer 1

The correct option is c: 1.5 cm.

Explanation:

The amplitude of the motion can be calculated using the following equation:

`A = (v)/(\omega)`

Where:

A: is the amplitude

v: is the speed = 15 cm/s

ω: is the angular speed

First, we need to find the angular speed:

`\omega = (2\pi)/(T)`

Where:

T is the period = 628 ms

`\omega = (2\pi)/(T) = (2\pi)/(0.628 s) = 10.01 rad/s`

Now we can find the amplitude:

`A = (v)/(\omega) = (15 cm/s)/(10.01 s) = 1.5 cm`

Therefore, the correct option is c: 1.5 cm.

I hope it helps you!