Question

7. A particle executes SHM on a line 8 cm long. Its KE and PE
will be equal when its distance from the mean position is
(a) 4 cm (b) 2 cm 6C) 212 cm (d) 2 cm​

Answer 1

Answer:

sorry I don't know the answer. I am extremely sorry

Answer 2

Answer:

C) 2√2 cm

Explanation:

For SHM, we know that

PE = $\frac{1}{2}kx^{2}$

KE= $\frac{1}{2}k(A^{2} -x^{2} )$

Here A= amplitude = 8/2 = 4 cm (distance from mean to extreme position)

Equating both, we get

$x^{2} = 4^{2} -x^{2}$

$2x^{2} = 16$

$x^{2} = 8$

$x = 2\sqrt{2}$

So the distance from mean position when KE and PE are equal is 2√2 cm