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.

Respuesta :

Answer:

The correct option is c: 1.5 cm.

Step-by-step explanation:

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

[tex] A = \frac{v}{\omega} [/tex]

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:

[tex] \omega = \frac{2\pi}{T} [/tex]

Where:

T is the period = 628 ms

[tex] \omega = \frac{2\pi}{T} = \frac{2\pi}{0.628 s} = 10.01 rad/s [/tex]

Now we can find the amplitude:

[tex] A = \frac{v}{\omega} = \frac{15 cm/s}{10.01 s} = 1.5 cm [/tex]

Therefore, the correct option is c: 1.5 cm.

I hope it helps you!