Grade 11 : Physics - Oscillations (Set 4)

Explanation: At $t = 0$, $x(0) = A \sin(\phi) = A/2 \implies \sin(\phi) = 1/2$. This gives $\phi = \pi/6$ or $\phi = 5\pi/6$. The velocity is $v(t) = A \omega \cos(\omega t + \phi)$, so at $t=0$, $v(0) = A \omega \cos(\phi)$. Since the particle moves in the negative x-direction, $v(0) < 0 \implies \cos(\phi) < 0$. Therefore, $\phi = 5\pi/6$.

Explanation: The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$, so $T \propto \sqrt{m}$. Taking ratios: $\frac{T_2}{T_1} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{2.00}{0.50}} = \sqrt{4} = 2$. Thus, $T_2 = 2 \times 1.00\text{ s} = 2.00\text{ s}$.

Explanation: The amplitude equals the circle's radius: $A = R = 4m$. Angular speed is $\omega = \frac{v}{R} = \frac{2\pi}{4} = \frac{\pi}{2}\text{ rad/s}$. The frequency is $f = \frac{\omega}{2\pi} = \frac{\pi/2}{2\pi} = 0.25\text{ Hz}$.

Explanation: $U = \frac{1}{2} k x^2$ and $K = \frac{1}{2} k (A^2 - x^2)$. Given $K = 3U \implies \frac{1}{2}k(A^2 - x^2) = 3 \left(\frac{1}{2}k x^2\right)$. Simplifying gives $A^2 - x^2 = 3x^2 \implies 4x^2 = A^2 \implies x = \pm \frac{A}{2}$.

Explanation: Effective acceleration due to gravity is $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$. The ratio of periods is $\frac{T}{T_0} = \sqrt{\frac{g}{g_{eff}}} = \sqrt{\frac{g}{4g/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Thus $T = 2.0 \times \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.73\text{ s}$.

Explanation: $x_2(t) = 4 \cos(\omega t) = 4 \sin(\omega t + \pi/2)$. The phase difference between $x_1$ and $x_2$ is $\phi = \pi/2$. Resultant amplitude is $A_{net} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\pi/2)} = \sqrt{3^2 + 4^2 + 0} = \sqrt{25} = 5\text{ units}$.

Explanation: In equal time intervals of 10 s, the amplitude decays by a constant factor of $\frac{4.0}{8.0} = 0.5$. Therefore, in the next 10 s (at $t = 20\text{ s}$), the amplitude becomes $4.0 \times 0.5 = 2.0cm$.

Explanation: For springs in parallel, equivalent stiffness is $k_{eq} = k_1 + k_2 = 100 + 300 = 400\text{ N/m}$. Angular frequency $\omega = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{400}{1.0}} = 20\text{ rad/s}$. Frequency $f = \frac{\omega}{2\pi} = \frac{20}{2\pi} = \frac{10}{\pi}\text{ Hz}$.

Explanation: Standard velocity-displacement relation in SHM is $\frac{x^2}{A^2} + \frac{v^2}{\omega^2 A^2} = 1$. Comparing gives $A^2 = 16 \implies A = 4m$ and $\omega^2 A^2 = 64 \implies v_{max} = 8m/s$. Then $\omega = 8/4 = 2\text{ rad/s}$. Period $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi\text{ s}$.

Explanation: Period of oscillation for liquid column in U-tube is $T = 2\pi \sqrt{\frac{h}{g}} = 2\pi \sqrt{\frac{0.2}{9.8}} = 2\pi \sqrt{\frac{1}{49}} = \frac{2\pi}{7} \approx 0.90\text{ s}$.

Explanation: For physical pendulum, $T = 2\pi \sqrt{\frac{I}{m g d}}$. Here $I = \frac{1}{3} m L^2$ and distance to center of mass $d = L/2$. So $T = 2\pi \sqrt{\frac{2L}{3g}} = 2\pi \sqrt{\frac{2(1.2)}{3(9.8)}} = 2\pi \sqrt{\frac{0.8}{9.8}} = 2\pi \times \frac{2}{7} = \frac{4\pi}{7} \approx 1.79\text{ s}$.

Explanation: Restoring force for displacement $x$ is $F = -\rho A g x$, so effective spring constant $k_{eq} = \rho A g = 1000 \times 10^{-3} \times 10 = 10\text{ N/m}$. Time period $T = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{0.5}{10}} = 2\pi \sqrt{\frac{1}{20}} = \frac{\pi}{\sqrt{5}} \approx 1.40\text{ s}$.

Explanation: The amplitude reaches a sharp peak (25 mm) when driving frequency $f_d = 20\text{ Hz}$. Resonance occurs when the driving frequency matches the natural frequency of the system, causing maximum energy transfer and maximum amplitude.

Explanation: Equation of motion starting from mean position is $x(t) = A \sin(\omega t)$. Setting $x = A/2 \implies A/2 = A \sin(\omega t) \implies \sin(\omega t) = 1/2$. Smallest positive angle is $\omega t = \pi/6$. Since $\omega = 2\pi/T$, we get $\frac{2\pi}{T} t = \frac{\pi}{6} \implies t = \frac{T}{12}$.

Explanation: Time period $T \propto \sqrt{L}$. Thus $\frac{T_1}{T_2} = \sqrt{\frac{1.0}{1.44}} = \frac{1}{1.2} = \frac{5}{6}$. Therefore $6 T_1 = 5 T_2$. After $N = 6$ oscillations of the shorter pendulum, the longer pendulum completes 5 oscillations, bringing them back in phase.

Explanation: In series: $k_s = k/2 \implies T_s = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}}$. In parallel: $k_p = 2k \implies T_p = 2\pi \sqrt{\frac{m}{2k}}$. Ratio $\frac{T_s}{T_p} = \frac{\sqrt{2}}{\sqrt{1/2}} = \sqrt{4} = 2$. So ratio is $2:1$.

Explanation: In SHM, $a = -\omega^2 x$. The slope of $a$ vs $x$ graph is $-\omega^2 = -100 \implies \omega = 10\text{ rad/s}$. Time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \approx 0.628\text{ s}$.

Explanation: Fractional change in period is $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta = \frac{1}{2} (2 \times 10^{-5}) (40 - 20) = 2 \times 10^{-4}$. Time lost per day (86400 seconds) is $\Delta t = \left(\frac{\Delta T}{T}\right) \times 86400 = 2 \times 10^{-4} \times 86400 = 17.28\text{ s}$.

Explanation: Frequencies are $f_1 = \frac{100\pi}{2\pi} = 50\text{ Hz}$ and $f_2 = \frac{104\pi}{2\pi} = 52\text{ Hz}$. The beat frequency is $f_{beat} = |f_2 - f_1| = 52 - 50 = 2\text{ Hz}$.

Explanation: Total mechanical energy in SHM is $E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} (1.0) (10)^2 (0.2)^2 = \frac{1}{2} \times 100 \times 0.04 = 2.0\text{ J}$.