Oscillations — NEET UG practice

1. The time period $T$ of a pendulum is the time for one complete oscillation. Since 30 oscillations take 60 s, dividing gives the time for one oscillation:

$$T = \frac{60}{30} = 2 \text{ s}$$

2. The relationship between the time period and the length of a simple pendulum is given by the standard formula:

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

3. Rearranging this formula to isolate $L$ (squaring both sides and solving) gives:

$$L = \frac{gT^2}{4\pi^2}$$

4. Substituting $g = 9.8 \text{ m/s}^2$, $T = 2\text{ s}$, and $\pi^2 = 9.8$:

$$L = \frac{9.8 \times (2)^2}{4 \times 9.8} = \frac{9.8 \times 4}{39.2} = 1 \text{ m}$$

Hence, the answer is D. $1 \text{ m}$.

1. For a particle in SHM, the velocity at any displacement $x$ from the mean position is given by:

$$v = \omega \sqrt{A^2 - x^2}$$

where $A$ is the amplitude and $\omega$ is the angular frequency.

2. Squaring both sides so that $v$ appears without a square root helps us compare it with the equation of a circle:

$$v^2 = \omega^2 (A^2 - x^2)$$

3. Rearranging this so that $x^2$ and the $v^2$ term are on the same side puts it in a recognisable geometric form:

$$\frac{v^2}{\omega^2} + x^2 = A^2$$

4. A circle of radius $A$ in the $x$-$v$ plane needs equal coefficients for $x^2$ and $v^2$. Comparing the equation above with $x^2 + v^2 = A^2$, this is only possible if:

$$\omega^2 = 1$$

5. Since for a spring-mass system $\omega^2 = \dfrac{k}{m}$, substitute $\omega^2 = 1$:

$$\frac{k}{m} = 1 \implies k = m$$

Hence, the answer is C. $k=m$.

1. The kinetic energy of any object is given by the standard formula:

$$K = \frac{1}{2}mv^2$$

2. For a pendulum bob in SHM, its speed at time $t$ is $v = A\omega \sin(\omega t + \phi)$... but since we want K.E. in terms of $\cos$ for comparison with the position curve, we can write the speed using the cosine form of velocity, so that substituting into the kinetic energy formula gives:

$$K = \frac{1}{2}mA^2\omega^2 \cos^2(\omega t + \phi)$$

3. Since $m$, $A$, and $\omega$ are all constants for a given pendulum, the kinetic energy varies only through the squared cosine term:

$$K \propto \cos^2(\omega t + \phi)$$

4. A $\cos^2$ function is always non-negative, oscillates between $0$ and a maximum value, and completes two humps per full time period $T$ (since $\cos^2\theta$ repeats twice as fast as $\cos\theta$).

Hence, the graph matching this double-humped, always-positive shape is option C.