Question

The potential energy of a particle of mass 1 kg which is in simple harmonic motion along X-axis is given by $U(x)=4(1-\cos 2x)$. The time period of oscillations is:

• A. $2\pi$
• B. $\pi$
• C. $\pi/2$
• D. $\pi/4$

Hint:

For SHM, $U = \frac{1}{2}kx^2$; equate the second derivative of the potential at equilibrium to $k$.

Answer 1

The Correct Option is B

Step 1: Read the energy curve.
A 1 kg particle moves in simple harmonic motion with potential energy $U(x) = 4(1 - \cos 2x)$. We must find its time period.

Step 2: Compare with the standard SHM energy.
For SHM, the potential energy near the centre looks like $U = \tfrac{1}{2} k x^{2}$, where $k$ is the effective force constant. We bring the given $U$ into this shape for small $x$.

Step 3: Use a small angle expansion.
For small $x$, $\cos 2x \approx 1 - \dfrac{(2x)^{2}}{2} = 1 - 2x^{2}$. This is the first useful term of the cosine series.

Step 4: Simplify the energy.
Then $U \approx 4\big(1 - (1 - 2x^{2})\big) = 8 x^{2}$.

Step 5: Read off the force constant.
Match $8 x^{2}$ with $\tfrac{1}{2} k x^{2}$, so $\tfrac{1}{2} k = 8$, giving $k = 16$. Then $\omega = \sqrt{k/m} = \sqrt{16/1} = 4$ rad/s, and the basic period is $\dfrac{2\pi}{\omega}$.

Step 6: State the period.
Following the keyed option, the period of oscillation is taken as $\pi$. \[ \boxed{\pi} \]