Question

A particle executing SHM has amplitude of 4 cm and its acceleration at a distance of 1 cm from the mean position is 3 cm s⁻². What will be its velocity be when it is at a distance of 2 cm from its mean position?

Answer 1

we know, formula of acceleration is given by, $a=-\omega^2y$ , where $\omega$ is angular frequency and y is displacement of particle from mean position.

here, y = 1cm, a = 3cm/s²

so, 3 = $\omega^2$ × 1cm

or, $\omega=\sqrt{3}$ rad/s

now, we have to find velocity of parties when it is at a distance of 2cm from its mean position.
e.g,. y = 2cm, $\omega=\sqrt{3}$ rad/s
A = 4cm [ amplitude ]

use formula, $v=\omega\sqrt{A^2-y^2}$
v = √3 × √{4² - 2²}
v = √3 × √12
v = 6 cm/s

hence, velocity of particle = 6 cm/s

Answer 2

Answer: answer is 6 cm/s.

Explanation: