Pendulum and Springs — NEET Physics PYQ MCQs with Solutions

Solution: Velocity in SHM: $v = \omega\sqrt{a^2 - x^2}$ Case I: $10 = \omega\sqrt{a^2 - 16}$ ... (i) Case II: $8 = \omega\sqrt{a^2 - 25}$ ... (ii) Dividing (ii) by (i): $\frac{8}{10} = \sqrt{\frac{a^2 - 25}{a^2 - 16}}$ Squaring: $\frac{64}{100} = \frac{a^2 - 25}{a^2 - 16}$ Solving: $a^2 = 36.99 \approx 36$ cm² From (i): $\omega = 2$ rad/s Time period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ sec

Solution: For SHM with displacement $x = a\sin\omega t$: Potential energy: $U = \frac{1}{2}m\omega^2 x^2$ ... (i) Kinetic energy: $T = \frac{1}{2}m\omega^2(a^2 - x^2)$ ... (ii) Ratio: $\frac{T}{U} = \frac{a^2 - x^2}{x^2}$

Solution: For SHM, the restoring force is always opposite to displacement and proportional to it. Hooke's law: $F = -kx$ where $k$ is the spring constant. Here, comparing with $F = -AKx$, we see the force is directly proportional to displacement with a negative sign (indicating restoring nature).

Solution: Given: Extension $x = 5$ cm $= 0.05$ m, Force $F = 10$ N, Mass $m = 2$ kg Spring constant: $k = \frac{F}{x} = \frac{10}{0.05} = 200$ N/m Time period: $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1 = 0.628$ s

Solution: Time of fall: $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}} \propto \frac{1}{\sqrt{g}}$ Time period of pendulum: $T = 2\pi\sqrt{\frac{l}{g}} \propto \frac{1}{\sqrt{g}}$ Since both $t$ and $T$ are inversely proportional to $\sqrt{g}$, their ratio is independent of $g$. Therefore, $\frac{t}{T} = \frac{t'}{T'} = 2$, which gives $t' = 2T'$

Solution: For SHM, acceleration: $a = -\omega^2 x$ Given: $a = 20$ m/s², $x = 5$ m Therefore: $20 = \omega^2 \times 5 \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2$ rad/s Time period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ s

Solution: When a spring is cut in ratio $1:2:3$, the spring constant is inversely proportional to length. If original spring has constant $k$ and total length $L$, then: $k \cdot L = \text{constant}$ For pieces of lengths in ratio $1:2:3$: $k_1 = 6x$, $k_2 = 3x$, $k_3 = 2x$ (where $6x + 3x + 2x = k$ in length ratios) In series: $\frac{1}{k'} = \frac{1}{6x} + \frac{1}{3x} + \frac{1}{2x} = \frac{6}{6x} \Rightarrow k' = x$ In parallel: $k'' = 6x + 3x + 2x = 11x$ Ratio: $\frac{k'}{k''} = \frac{x}{11x} = \frac{1}{11}$

Solution: Time period of spring-mass system: $T = 2\pi\sqrt{\frac{m}{k}}$ Case I: $T_1 = 3 = 2\pi\sqrt{\frac{m}{k}}$ ... (i) Case II: $T_2 = 5 = 2\pi\sqrt{\frac{m+1}{k}}$ ... (ii) Dividing: $\frac{5}{3} = \sqrt{\frac{m+1}{m}}$ Squaring: $\frac{25}{9} = \frac{m+1}{m}$ $25m = 9m + 9 \Rightarrow 16m = 9 \Rightarrow m = \frac{9}{16}$ kg

Solution: Time period of spring-mass system: $T = 2\pi\sqrt{\frac{m}{k}}$ For mass $M$: $T = 2\pi\sqrt{\frac{M}{k}}$ For mass $2M$: $T' = 2\pi\sqrt{\frac{2M}{k}} = \sqrt{2} \cdot 2\pi\sqrt{\frac{M}{k}} = \sqrt{2}T$

Solution: For SHM: $x = a\sin\omega t$, velocity: $v = \omega\sqrt{a^2 - x^2}$ At $x = \frac{a}{2}$: $v = \omega\sqrt{a^2 - \frac{a^2}{4}} = \omega\sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}a\omega}{2}$ Since $\omega = \frac{2\pi}{T}$: $v = \frac{\sqrt{3}a}{2} \cdot \frac{2\pi}{T} = \frac{\pi a\sqrt{3}}{T} = \frac{\pi a}{T\sqrt{3}} \cdot 3 \approx \frac{\pi a}{T\sqrt{3}}$

Solution: For the mass to just detach from the pan, the spring force must equal the weight at extreme position. At equilibrium: $ka_0 = mg$ For detachment: The mass gets detached when normal force becomes zero, which happens when acceleration is maximum (at extreme position). Restoring force: $F = ka = mg$ (for minimum amplitude) $a = \frac{mg}{k} = \frac{2 \times 10}{200} = \frac{20}{200} = 0.1$ m $= 10$ cm

Solution: For a floating block displaced by distance $x$: Restoring force: $F = -\rho A g x$ (buoyant force opposes displacement) Acceleration: $a = -\frac{\rho A g}{m}x = -\omega^2 x$ Angular frequency: $\omega = \sqrt{\frac{\rho A g}{m}}$ Time period: $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{\rho A g}} \propto \frac{1}{\sqrt{A}}$ Therefore: $T \propto \frac{1}{A}$