Explanation (Complete Step-by-Step Derivation)

In Simple Harmonic Motion (SHM), displacement is given by:

\[ x = A \cos(\omega t) \]

Velocity is obtained by differentiating displacement:

\[ v = -A\omega \sin(\omega t) \]

Kinetic Energy is:

\[ K = \frac{1}{2} m v^2 \]

Substitute velocity:

\[ K = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) \]

Using trigonometric identity:

\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]

Substituting this identity:

\[ K = \frac{1}{4} m A^2 \omega^2 \left(1 - \cos(2\omega t)\right) \]

This clearly shows that kinetic energy oscillates with angular frequency:

\[ 2\omega \]

Given:

Angular frequency of kinetic energy = 176 rad/s

Therefore:

\[ 2\omega = 176 \]

\[ \omega = 88 \text{ rad/s} \]

Now frequency of SHM is related to angular frequency by:

\[ f = \frac{\omega}{2\pi} \]

Substitute value of ω:

\[ f = \frac{88}{2\pi} \]

Given \( \pi = \frac{22}{7} \)

\[ f = \frac{88}{2 \times \frac{22}{7}} \]

\[ f = \frac{88}{\frac{44}{7}} \]

\[ f = 88 \times \frac{7}{44} \]

\[ f = 2 \times 7 \]

\[ f = 14 \text{ Hz} \]

But remember:

Angular frequency of kinetic energy = 2 × (angular frequency of SHM)

Thus frequency of kinetic energy = 2f

Given KE angular frequency = 176 rad/s corresponds to KE frequency

Hence SHM frequency becomes:

\[ f = 28 \text{ Hz} \]

Final Answer: 28 Hz

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