Question

58. A particle is describing SHM with amplitude 'a'.
When the potential energy of particle is one fourth
of the maximum energy during oscillation, then its
displacement from mean position will be:​

Answer 1

Answer:

Given:

Amplitude = a

Potential energy is ¼ of total energy during oscillation at a distance"x" from mean position.

To find:

Value of "x"

Calculation:

∴ Potential energy at x = ¼(max energy)

=> ½mω²x² = ¼ [ ½mω²a² ]

=> x² = ¼ a²

=> x = ½ (a)

So final answer is :

$\boxed{ \boxed{ \red{displacement = \dfrac{a}{2}}}}$

Answer 2

ANSWER:-

GIVEN:-

AMPLITUDE =a

P. E=¼

DISTANCE =X.

FIND:-

VALUE OF X.

SOLUTION :-

P. E at x=¼(M. E)

$\implies$½mw²x²=¼(½mw²a²)

.°.$\implies$x²=¼a²

.°.$\implies$ x=(√¼a²)

.°.$\implies$ x=½ a.