Question

The kinetic energy of a particle performing S.H.M. is $\dfrac{1{n}$ times its potential energy. If the amplitude of S.H.M. is $A$, then the displacement of the particle will be

• A. $\sqrt{\dfrac{nA^2}{n+1}}$
• B. $\dfrac{A}{n}$
• C. $nA$
• D. $\sqrt{\dfrac{(n+1)A^2}{n}}$

Hint:

In SHM, kinetic energy decreases as potential energy increases, but their sum always remains constant.

Answer 1

The Correct Option is A

Step 1: Write expressions for energies in SHM.
Potential energy at displacement $x$: \[ PE = \dfrac{1}{2}k x^2 \] Kinetic energy: \[ KE = \dfrac{1}{2}k(A^2 - x^2) \] Step 2: Apply given condition.
\[ KE = \dfrac{1}{n} PE \] Step 3: Substitute energy expressions.
\[ \dfrac{1}{2}k(A^2 - x^2) = \dfrac{1}{n}\left(\dfrac{1}{2}k x^2\right) \] Step 4: Simplify the equation.
\[ A^2 - x^2 = \dfrac{x^2}{n} \] \[ A^2 = x^2\left(1 + \dfrac{1}{n}\right) \] Step 5: Solve for displacement $x$.
\[ x^2 = \dfrac{nA^2}{n+1} \] \[ x = \sqrt{\dfrac{nA^2}{n+1}} \]