JEE Main 17 March 2021 Shift 1 Solved Paper

Potential energy of SHM=‌

1

2

kx2=‌

1

2

mω2x2
where m= mass of particle,
ω= angular velocity
and x= displacement.
Kinetic energy of SHM=‌

1

2

m(ω√A2−x2)2
Here, A is the amplitude of SHM.
According to question,
Potential energy of SHM= Kinetic energy of SHM
‌

1

2

mω2x2=‌

1

2

m(ω√A2−x2)2
⇒‌‌2x2=A2⇒x2=‌

A2

2

⇒‌‌x=±√‌

A2

2

⇒x=±‌

A

√2