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JAMB Physics 2017 Paper

Question : 36

Total: 49

What is the speed of a body vibrating at 50 cyclic per second

Solution:

Concept:

Angular speed for periodic vibration is calculated from frequency using

\omega = 2\pi f

Explanation:

One complete cycle equals

2\pi

radians.

The body vibrates at

50

cycles per second, so

f = 50

Hz.

Thus, angular speed

\omega = 2\pi f

Substitute

f = 50

\omega = 2\pi \times 50

This gives

\omega = 100\pi

rad/s.

Answer:

100\pi

rad s

^{-1}

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