Oscillations

(a) Swimmer trip: Periodic. The motion of the swimmer repeats itself over regular intervals of time (back and forth).

(b) Suspended bar magnet: Periodic. When displaced, it oscillates about its North-South equilibrium position due to the earth’s magnetic field.

(c) Hydrogen molecule rotation: Periodic. A molecule rotating about its centre of mass repeats its orientation at regular intervals.

(d) Arrow released from bow: Non-periodic. The arrow moves in one direction and does not return to its starting point to repeat the motion.

(a) Earth’s Rotation: Periodic but NOT SHM. It repeats regularly but there is no to-and-fro oscillation about a mean position.

(b) Oscillating Mercury Column: SHM. The restoring force (weight of the raised column) is directly proportional to displacement ($F \propto -y$).

(c) Ball in Curved Bowl: SHM. For small displacements, the restoring force is proportional to displacement.

(d) Polyatomic Molecule Vibrations: Periodic but NOT SHM. These are superpositions of multiple normal modes, resulting in complex periodic motion, not simple harmonic.

(Based on standard NCERT Figure 13.18)

(a) Sine Curve: Periodic. It repeats every $T=2$ s.

(b) Alternating Curve (Sine + Cosine): Periodic. Repeats every $T=2$ s.

(c) Non-repeating Curve: Non-Periodic. The pattern does not repeat.

(d) Logarithmic/Exponential Curve: Non-Periodic. It never repeats its values.

(a) $\sin \omega t – \cos \omega t$: SHM.
Can be written as $\sqrt{2} \sin(\omega t – \pi/4)$. Period $T = 2\pi/\omega$.

(b) $\sin^3 \omega t$: Periodic but NOT SHM.
$\sin^3 \theta = (3\sin \theta – \sin 3\theta)/4$. It is a sum of two SHMs with different frequencies. Period $T = 2\pi/\omega$.

(c) $3 \cos (\pi/4 – 2\omega t)$: SHM.
It is a cosine function. Angular frequency is $2\omega$. Period $T = 2\pi/2\omega = \pi/\omega$.

(d) $\cos \omega t + \cos 3\omega t + \cos 5\omega t$: Periodic but NOT SHM.
Superposition of periodic functions. Period is LCM of individual periods ($2\pi/\omega$).

(e) $\exp(-\omega^2 t^2)$: Non-Periodic.
It is an exponential decay function that never repeats.

(f) $1 + \omega t + \omega^2 t^2$: Non-Periodic.
It is a polynomial function that increases monotonically.

Range $AB = 10 \text{ cm}$. Midpoint $O$ is origin. $A = -5, B = +5$.
Acceleration and Force are always directed towards equilibrium $O$.

(a) At end A (-5 cm): Velocity $v=0$. Acc/Force towards $O$ (+ve).
$v=0, a>0, F>0$.

(b) At end B (+5 cm): Velocity $v=0$. Acc/Force towards $O$ (-ve).
$v=0, a<0, F<0$.

(c) Midpoint towards A: Moving Left ($v<0$). At $x=0$, Force/Acc is zero.
$v<0, a=0, F=0$.

(d) 2 cm from B going to A: Pos $x=+3$. Moving Left ($v<0$). Force towards $O$ ($a<0, F<0$).
$v<0, a<0, F<0$.

(e) 3 cm from A going to B: Pos $x=-2$. Moving Right ($v>0$). Force towards $O$ ($a>0, F>0$).
$v>0, a>0, F>0$.

(f) 4 cm from B going to A: Pos $x=+1$. Moving Left ($v<0$). Force towards $O$ ($a<0, F<0$).
$v<0, a<0, F<0$.

Condition for SHM: Acceleration is proportional to negative displacement ($a \propto -x$).

(c) $a = -10x$: This is the only relation that satisfies the SHM condition.

(a) is linear but positive (unstable). (b) and (d) depend on higher powers of $x$.

Given: $x(t) = A \cos(\omega t + \phi)$. At $t=0$, $x=1$ cm, $v=\omega$ cm/s.
$v(t) = -A\omega \sin(\omega t + \phi)$.

At $t=0$:
1. $1 = A \cos \phi$
2. $\omega = -A\omega \sin \phi \Rightarrow \sin \phi = -1$.

From (2), $\phi = -\pi/2$ or $3\pi/2$.
From (1), $A \cos(-\pi/2) = 0 \neq 1$. Contradiction? No, magnitude of amplitude. Squaring and adding: $x^2 + (v/\omega)^2 = A^2(\cos^2 + \sin^2)$.
$1^2 + (1)^2 = A^2 \Rightarrow A = \sqrt{2}$ cm.
$\cos \phi = 1/\sqrt{2}, \sin \phi = -1/\sqrt{2}$. This corresponds to $\phi = -\pi/4$ or $7\pi/4$.

Wait, earlier velocity calculation assumed $v=\omega$. If $v=\omega$, then $\sin \phi = -1/\sqrt{2}$ isn’t consistent with just $\sin \phi=-1$. Let’s re-read: “Initial velocity is $\omega$”. $v = -A\omega \sin \phi = \omega \Rightarrow A \sin \phi = -1$. $x = A \cos \phi = 1$. $A^2 = (1)^2 + (-1)^2 = 2 \Rightarrow \mathbf{A = \sqrt{2} \text{ cm}}$. $\tan \phi = -1$. Since $\cos>0, \sin<0$, $\phi$ is in 4th quadrant. $\mathbf{\phi = -\pi/4 \text{ or } 7\pi/4}$.

Max Mass $M = 50 \text{ kg}$. Scale Length $L = 0.2 \text{ m}$.
Spring Constant $k = \frac{Mg}{L} = \frac{50 \times 9.8}{0.2} = 2450 \text{ N/m}$.

Period $T = 0.6 \text{ s}$. Formula: $T = 2\pi \sqrt{\frac{m}{k}}$.
$m = \frac{k T^2}{4\pi^2} = \frac{2450 \times (0.6)^2}{4 \times (3.14)^2} \approx 22.36 \text{ kg}$.

Weight of body $W = mg = 22.36 \times 9.8 \approx \mathbf{219.1 \text{ N}}$.

Given: $k = 1200 \text{ N/m}$, $m = 3 \text{ kg}$, Amplitude $A = 2 \text{ cm} = 0.02 \text{ m}$.

(i) Frequency:
Angular freq $\omega = \sqrt{k/m} = \sqrt{1200/3} = \sqrt{400} = 20 \text{ rad/s}$.
Frequency $\nu = \frac{\omega}{2\pi} = \frac{20}{2\pi} \approx \mathbf{3.18 \text{ Hz}}$.

(ii) Max Acceleration:
$a_{max} = \omega^2 A = (20)^2 \times 0.02 = 400 \times 0.02 = \mathbf{8 \text{ m/s}^2}$.

(iii) Max Speed:
$v_{max} = \omega A = 20 \times 0.02 = \mathbf{0.4 \text{ m/s}}$.

Parameters: $A = 2 \text{ cm}$, $\omega = 20 \text{ s}^{-1}$.

(a) Mean Position ($x=0$ at $t=0$):
If velocity is positive (towards right): $x = A \sin(\omega t)$.
Equation: $\mathbf{x = 2 \sin(20t)}$.

(b) Max Stretched ($x=+A$ at $t=0$):
Equation: $\mathbf{x = 2 \cos(20t)}$.

(c) Max Compressed ($x=-A$ at $t=0$):
Equation: $\mathbf{x = -2 \cos(20t)}$ or $2 \cos(20t + \pi)$.

(a) Particle P starts at angle $225^\circ$ ($5\pi/4$) or $-135^\circ$. Radius $R=3$ cm. Period $T=2$ s. Rotation anticlockwise.
$\omega = 2\pi/T = \pi$.
$x$-projection: $x = R \cos(\omega t + \phi)$.
Equation: $\mathbf{x = 3 \cos(\pi t + 3\pi/2)}$ (if starting from Y-axis? Need diagram specifics. Usually angle is from X-axis). Assuming standard convention.

(b) Similar logic applies based on the specific starting angle provided in the textbook diagram.

(a) $x = -2 \sin(3t + \pi/3)$:
Convert to cosine: $x = 2 \cos(3t + \pi/3 + \pi/2) = 2 \cos(3t + 5\pi/6)$.
Circle Radius $R=2$ cm. Start Angle $150^\circ$.

(b) $x = \cos(\pi/6 – t)$:
Cos is even function: $x = \cos(t – \pi/6)$.
Radius $R=1$ cm. Start Angle $-30^\circ$.

(c) $x = 3 \sin(2\pi t + \pi/4)$:
$x = 3 \cos(2\pi t + \pi/4 – \pi/2) = 3 \cos(2\pi t – \pi/4)$.
Radius $R=3$ cm. Start Angle $-45^\circ$.

(d) $x = 2 \cos \pi t$:
Radius $R=2$ cm. Start Angle $0^\circ$.

(a) Maximum Extension:
Case (a): Wall and Mass. Force $F$. Extension $x = F/k$.
Case (b): Two masses pulling with $F$ each. The tension in the spring is $F$ throughout (Action-Reaction). Extension $x = F/k$.
Result: Extension is the same in both cases.

(b) Period of Oscillation:
Case (a): Standard spring-mass. $T = 2\pi \sqrt{m/k}$.
Case (b): Two body oscillator. Reduced mass $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$.
Period $T = 2\pi \sqrt{\mu/k} = 2\pi \sqrt{m/2k}$.
Alternatively, consider the center of the spring fixed (node). Each half has length $L/2$ and constant $2k$.
$T = 2\pi \sqrt{m/2k}$.

Stroke $= 2 \times \text{Amplitude} = 1.0 \text{ m} \Rightarrow A = 0.5 \text{ m}$.
Angular frequency $\omega = 200 \text{ rad/min} = \frac{200}{60} \text{ rad/s} = \frac{10}{3} \text{ rad/s}$.

Max Speed $v_{max} = A\omega = 0.5 \times \frac{10}{3} = \frac{5}{3} \approx \mathbf{1.67 \text{ m/s}}$.

Time period formula: $T = 2\pi \sqrt{\frac{l}{g}}$. Hence $T \propto \frac{1}{\sqrt{g}}$.
$\frac{T_m}{T_e} = \sqrt{\frac{g_e}{g_m}} = \sqrt{\frac{9.8}{1.7}} = \sqrt{5.76} \approx 2.4$.

$T_m = 2.4 \times T_e = 2.4 \times 3.5 = \mathbf{8.4 \text{ s}}$.

The bob experiences two accelerations:
1. Gravity $g$ (downwards).
2. Centripetal acceleration $a_c = v^2/R$ (radial/horizontal).

Effective gravity $g_{eff} = \sqrt{g^2 + (v^2/R)^2}$.

Time Period $T = 2\pi \sqrt{\frac{l}{g_{eff}}} = \mathbf{2\pi \sqrt{\frac{l}{\sqrt{g^2 + v^4/R^2}}}}$.

Equilibrium: Weight of cork = Buoyant force.
$Ah\rho g = A y_0 \rho_l g$ (where $y_0$ is submerged depth).
$\Rightarrow h\rho = y_0 \rho_l$.

Displacement $y$:
Extra Buoyant Force = Restoring Force.
$F = -(A y \rho_l g)$.
Acceleration $a = \frac{F}{m} = \frac{-A \rho_l g y}{A h \rho} = -\frac{\rho_l g}{h \rho} y$.
Comparing with $a = -\omega^2 y$: $\omega^2 = \frac{\rho_l g}{h \rho}$.
Period $T = 2\pi \sqrt{\frac{h\rho}{\rho_l g}}$.

Total length $L$. Mass $m = A L \rho$.
Displace one side by $y$ downwards. The other side rises by $y$.
Level difference = $2y$.
Restoring Force (Weight of excess column) $F = -(2y) A \rho g$.

Acceleration $a = F/m = \frac{-2A\rho g y}{A L \rho} = -\frac{2g}{L} y$.
This is SHM ($a \propto -y$).
$\omega^2 = \frac{2g}{L} \Rightarrow \omega = \sqrt{\frac{2g}{L}}$.
Period $T = 2\pi \sqrt{\frac{L}{2g}}$.