Question

A particle performing S.H.M. has velocity of 10 cm/s when it crosses the mean position. If

the amplitude of oscillation is 2cm, find the velocity when it is mid-way between mean

and extreme positions. ​

Answer 1

velocity of particle, v = 10cm/s when it crosses the mean position.

amplitude of oscillation = 2cm

you know as well,

velocity of particle is given by, $v=\omega\sqrt{A^2-x^2}$

at mean position, x = 0

so, velocity of particle becomes ,v = $\omega A$

e.g., 10cm/s = $\omega$ × 2cm

or, $\omega$ = 5 rad/s

displacement of particle from mean position when it is located at midway between the mean and extreme position is x = A/2 = 2cm/2 = 1cm

now, velocity of particle, v = $\omega\sqrt{A^2-x^2}$

= 5 × √{2² - 1²} = 8.66 cm/s

hence, answer is 8.66 cm/s

Answer 2

Answer: correct answer is 5√3 i.e. 8.66 cm/s.

Explanation: