Grade 11 : Physics - Oscillations (Set 3)

Explanation:
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi\text{ rad/s}$.
General equation: $x(t) = A \sin(\omega t + \phi)$.
At $t = 0$, $x(0) = 0.05 \sin(\phi) = 0.025 \implies \sin(\phi) = 0.5$.
Possible phases are $\phi = \pi/6$ or $\phi = 5\pi/6$.
Velocity $v(t) = A\omega \cos(\omega t + \phi)$. At $t=0$, $v(0) = A\omega \cos(\phi) < 0$ (moving in negative direction).
Since $\cos(5\pi/6) = -\frac{\sqrt{3}}{2} < 0$, we select $\phi = 5\pi/6$.
Thus, $x(t) = 0.05 \sin(\pi t + 5\pi/6)$.

Explanation:
Total mechanical energy $E = \frac{1}{2} m \omega^2 A^2$.
Kinetic energy $K = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
For $x = \frac{A}{2}$, $K = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) = \frac{3}{4} \left(\frac{1}{2} m \omega^2 A^2\right) = \frac{3}{4} E$.
Therefore, the ratio $\frac{K}{E} = \frac{3}{4}$.

Explanation:
We know $v_{\text{max}} = \omega A$ and $a_{\text{max}} = \omega^2 A$.
Dividing maximum acceleration by maximum speed:
$\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{36}{12} = 3\text{ rad/s}$.
Frequency $f = \frac{\omega}{2\pi} = \frac{3}{2\pi}\text{ Hz}$.

Explanation:
Rewrite $x_2 = 4 \sin\left(\omega t + \frac{\pi}{2}\right)$.
The phase difference between the two SHMs is $\Delta \phi = \frac{\pi}{2}$.
Resultant amplitude $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}$.
Since $\cos(\pi/2) = 0$, $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.

Explanation:
Initial period $T_0 = 2\pi \sqrt{\frac{L}{g}}$.
When accelerating upward at $a = g/3$, effective acceleration due to gravity is:
$g_{\text{eff}} = g + a = g + \frac{g}{3} = \frac{4g}{3}$.
New period $T' = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{L}{4g/3}} = \sqrt{\frac{3}{4}} \cdot 2\pi \sqrt{\frac{L}{g}} = \frac{\sqrt{3}}{2} T_0$.

Explanation:
For two identical springs connected in parallel, the equivalent spring constant is:
$k_{\text{eq}} = k_1 + k_2 = k + k = 2k$.
The time period of SHM is $T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} = 2\pi \sqrt{\frac{m}{2k}}$.

Explanation:
For springs in series, equivalent constant $k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$.
Frequency $f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eq}}}{m}} = \frac{1}{2\pi} \sqrt{\frac{k_1 k_2}{m(k_1 + k_2)}}$.

Explanation:
Apparent weight in liquid = $m g_{\text{eff}} = m g - \text{Buoyant force}$.
$\rho V g_{\text{eff}} = \rho V g - \sigma V g \implies g_{\text{eff}} = g \left(1 - \frac{\sigma}{\rho}\right)$.
Since $T \propto \frac{1}{\sqrt{g_{\text{eff}}}}$, we have:
$T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} = \frac{T_0}{\sqrt{1 - \frac{\sigma}{\rho}}}$.

Explanation:
Potential energy $U = \frac{1}{2} k x^2$.
$0.4 = \frac{1}{2} k (0.1)^2 = 0.005 k \implies k = \frac{0.4}{0.005} = 80\text{ N/m}$.
Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{80}{0.2}} = \sqrt{400} = 20\text{ rad/s}$.

Explanation:
Displacement: $x(t) = A \sin(\omega t)$.
Velocity: $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin\left(\omega t + \frac{\pi}{2}\right)$.
Hence, velocity leads displacement by a phase angle of $\frac{\pi}{2}\text{ rad}$ ($90^\circ$).

Explanation:
Time period of spring-mass system is $T = 2\pi \sqrt{\frac{m}{k}}$.
Thus $T \propto \sqrt{m}$.
$\frac{T_2}{T_1} = \sqrt{\frac{4m}{m}} = 2$.
$T_2 = 2 T_1 = 2 \times 2.0\text{ s} = 4.0\text{ s}$.

Explanation:
Amplitude equation for damped oscillator: $A(t) = A_0 e^{-\gamma t / 2}$.
For energy to halve: $e^{-\gamma t_1} = 1/2 \implies \gamma t_1 = \ln 2$.
For amplitude to halve: $e^{-\gamma t_2 / 2} = 1/2 \implies \frac{\gamma t_2}{2} = \ln 2$.
Equating both gives $\frac{\gamma t_2}{2} = \gamma t_1 \implies t_2 = 2 t_1$.

Explanation:
For forced oscillations with weak damping, the amplitude is given by $A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + (b\omega_d/m)^2}}$.
The amplitude reaches its peak (resonance) when the driving frequency matches the natural frequency of the oscillator, i.e., $\omega_d = \omega_0$.

Explanation:
When liquid level is displaced by $y$, restoring force $F = -(2y A \rho) g$.
Equation of motion: $(m) \ddot{y} + (2 A \rho g) y = 0 \implies (A L \rho) \ddot{y} + 2 A \rho g y = 0$.
$\ddot{y} + \frac{2g}{L} y = 0$.
Hence $\omega = \sqrt{\frac{2g}{L}}$, so time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{2g}}$.

Explanation:
Period $T \propto \sqrt{L} \implies L \propto T^2$.
New period $T' = 1.10 T$.
New length $L' = (1.10)^2 L = 1.21 L$.
Percentage increase in length $= \frac{L' - L}{L} \times 100% = (1.21 - 1) \times 100% = 21%$.

Explanation:
Instantaneous kinetic energy $K(t) = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t)$.
Average value of $\cos^2(\omega t)$ over a complete period is $\frac{1}{2}$.
Therefore, average kinetic energy $\langle K \rangle = \frac{1}{2} m \omega^2 A^2 \times \frac{1}{2} = \frac{1}{4} m \omega^2 A^2$.

Explanation:
Gravitational force at radial distance $r \le R$ is $F = -\frac{G M m r}{R^3} = -m \left(\frac{g}{R}\right) r$.
This is a restoring force of SHM form $F = -k r$ with $k = \frac{m g}{R}$.
Time period $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{mg/R}} = 2\pi \sqrt{\frac{R}{g}}$.

Explanation:
The force constant of a spring is inversely proportional to its length ($k \propto \frac{1}{L}$).
When the spring is cut in half, its length becomes $L' = \frac{L}{2}$.
Therefore, the new spring constant becomes $k' = 2k$.

Explanation:
Spring constant is inversely proportional to length ($k \cdot L = \text{constant}$).
Let total length be $3x$. Shorter piece $L_1 = x$, longer piece $L_2 = 2x$.
$k_1 \cdot x = k \cdot 3x \implies k_1 = 3k$.
$k_2 \cdot 2x = k \cdot 3x \implies k_2 = \frac{3}{2} k$.
Ratio $\frac{k_1}{k_2} = \frac{3k}{1.5k} = 2$.

Explanation:
Velocity in SHM is given by $v = \omega \sqrt{A^2 - x^2}$.
Substitute $x = \frac{A}{\sqrt{2}}$:
$v = \omega \sqrt{A^2 - \frac{A^2}{2}} = \omega \sqrt{\frac{A^2}{2}} = \frac{\omega A}{\sqrt{2}}$.
Since $v_{\text{max}} = \omega A$, we get $v = \frac{v_{\text{max}}}{\sqrt{2}}$.