Simple Harmonic Motion — JEE Main practice

1. In SHM, the kinetic energy oscillates at twice the angular frequency of the displacement, so the given value must first be divided by 2 to get the oscillator's actual angular frequency. Given $2\omega = 176$ rad/s,

$$\omega = \frac{176}{2} = 88\ \text{rad/s}$$

2. Convert angular frequency to ordinary frequency using $\omega = 2\pi f$. Therefore,

$$f = \frac{\omega}{2\pi} = \frac{88}{2\times\frac{22}{7}} = \frac{88}{\frac{44}{7}}$$

Hence,

$$f = 88\times\frac{7}{44} = 14\ \text{Hz}$$

Hence, the frequency of the oscillator is 14 Hz, so the answer is option D.

1. Start from the standard pendulum time period formula, since the graphs plot a rearranged version of this relation. Given

$$T = 2\pi\sqrt{\frac{L}{g}}$$

2. Square both sides and rearrange to express $1/T^2$ in terms of $L$, since that is the quantity plotted on the graphs. Therefore,

$$T^2 = \frac{4\pi^2 L}{g} \quad\Rightarrow\quad \frac{1}{T^2} = \left(\frac{g}{4\pi^2}\right)\frac{1}{L}$$

3. Recognize the form of this equation to identify the shape of the graph. Letting $k=\dfrac{g}{4\pi^2}$ (a positive constant) and $y=\frac{1}{T^2}$, $x=L$,

$$xy = k$$

This is the equation of a rectangular hyperbola: $y\to0$ as $x\to\infty$, and $y\to\infty$ as $x\to0$, so both axes act as asymptotes.

Hence, the correct graph is the rectangular hyperbola described above, so the answer is option A.

1. The time period of a simple pendulum depends on its length, so we use the standard pendulum formula to relate the two quantities. Given

$$T = 2\pi\sqrt{\frac{L}{g}} \quad\Rightarrow\quad T^2 \propto L$$

2. Find the time period in the first case from the given number of oscillations and total time. Given $L_1 = 30$ cm, $N_1 = 20$ oscillations in $t_1 = 10$ s,

$$T_1 = \frac{10}{20} = 0.5\ \text{s}$$

3. Find the time period in the second case the same way. Given $N_2 = 40$ oscillations in the same $t_2 = 10$ s,

$$T_2 = \frac{10}{40} = 0.25\ \text{s}$$

4. Use the proportionality $L\propto T^2$ to relate the two lengths and solve for the new length $L_2$. Therefore,

$$\frac{L_2}{L_1} = \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{0.25}{0.5}\right)^2 = \frac14$$

Hence,

$$L_2 = \frac{30}{4} = 7.5\ \text{cm}$$

Hence, the required length of the string is 7.5 cm, so the answer is option A.