Question

A particle executes SHM in a line
4 cm long. Its velocity when
passing through the centre of line is 12 cm/s. The period will b
(a) 2.047 s
(b) 1.047 s
(c) 3.047 s
(d) 0.047 s​

Answer 1

The time period will be 1.047 seconds.

Explanation:

We are given:

Length of line = Distance between extreme positions of oscillation = 4 cm

Amplitude "a" = 2 cm

V(max) = 12 cm / s

V(max) = ωa = 2πTa

T = 2πa.V(max)

T = 2 x 3.14 x 212

T = 1.047 seconds

Thus the time period will be 1.047 seconds.

Also learn more

Three pendulums A, B and C have frequency 05 Hz,  0.05 Hz and 0.005 Hz respectively. The number of  oscillations completed by pendulum A and C in a  time period equal to time taken by pendulum B to  complete 200 oscillations are respectively ?

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Answer 2

Explanation:

amplitude a= half of distance between two points