Question

Assertion (A): In S.H.M., kinetic and potential energies become equal when the distance is \[ \frac{1}{\sqrt{2}} \] times the amplitude. Reason (R): The potential energy of a particle executing S.H.M. is periodic with time and is maximum at the extreme displacement.

• (A) and (R) are true. (R) is the correct explanation of (A)
• (A) and (R) are true. (R) is not the correct explanation of (A)
• (A) is true, but (R) is false
• (A) is false, but (R) is true

Hint:

For S.H.M., \[ U=\frac{1}{2}kx^2, \qquad K=\frac{1}{2}k(A^2-x^2). \] Whenever \[ K=U, \] the displacement is \[ x=\frac{A}{\sqrt2}. \] This is a very common result in S.H.M. problems.

Answer 1

The Correct Option is B

Step 1: Verify Assertion — KE = PE condition.
\( KE = \frac{1}{2}m\omega^2(A^2 - x^2) \), \( PE = \frac{1}{2}m\omega^2 x^2 \). Setting equal: \( A^2 - x^2 = x^2 \Rightarrow x = \frac{A}{\sqrt{2}} \). Assertion is TRUE.

Step 2: Evaluate Reason.
PE is indeed periodic (with period T/2) and maximum at extreme positions. This statement is TRUE but it does not logically explain why KE = PE at \( x = A/\sqrt{2} \). So R is not the correct explanation of A.

\[ \boxed{\text{Both true; R not correct explanation of A}} \]