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PhysicsSimple Harmonic Motion
Two masses $m_A$ and $m_B$ are attached to two different springs $k_1$ and $k_2$ respectively. If the maximum velocities of both masses are same during their oscillations, find the ratio of their amplitudes $A_Q / A_P$:
Options
1
$\sqrt{k_2/k_1}$
2
$\sqrt{k_1/k_2}$
3
$k_1/k_2$
4
$k_2/k_1$
Correct Answer
$\sqrt{k_1/k_2}$
Solution
1

Maximum velocity in SHM:

$$v_{max} = A\omega = A\sqrt{\frac{k}{m}}$$

For P (spring $k_1$): $v_P = A_P\sqrt{k_1/m}$

For Q (spring $k_2$): $v_Q = A_Q\sqrt{k_2/m}$

2

Given $v_P = v_Q$ (same maximum velocity):

$$A_P\sqrt{\frac{k_1}{m}} = A_Q\sqrt{\frac{k_2}{m}}$$$$\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$$
$v_{max} = A\omega = A\sqrt{k/m}$
Same $v_{max}$: $A_Q/A_P = \sqrt{k_1/k_2}$
Theory: Simple Harmonic Motion
1. Simple Harmonic Motion

SHM: $x = A\sin(\omega t + \phi)$. Restoring force: $F = -kx$. Acceleration: $a = -\omega^2 x$. Angular frequency: $\omega = \sqrt{k/m}$. Period: $T = 2\pi\sqrt{m/k}$. Energy: $E = \frac{1}{2}kA^2$ (constant). $KE = \frac{1}{2}m\omega^2(A^2-x^2)$. $PE = \frac{1}{2}kx^2$. Velocity: $v = \omega\sqrt{A^2-x^2}$. Maximum velocity at $x=0$: $v_{max} = \omega A = A\sqrt{k/m}$.

2. Spring-Mass System

Horizontal spring: $T = 2\pi\sqrt{m/k}$. Vertical spring: $T = 2\pi\sqrt{m/k}$ (same! gravity only shifts equilibrium). Springs in series: $1/k_{eff} = 1/k_1 + 1/k_2$. Springs in parallel: $k_{eff} = k_1 + k_2$. If mass hung from two springs in series: equivalent to one spring of effective $k$. Springs connected to fixed wall in parallel: $k_{eff}$ adds. For spring cut into $n$ equal pieces: each piece has $k_{new} = nk$ (shorter spring is stiffer).

3. Simple Pendulum

$T = 2\pi\sqrt{L/g}$. Valid for small oscillations ($\theta < 5°$). Independent of mass and amplitude (for small angles). $T \propto \sqrt{L}$: doubling length multiplies $T$ by $\sqrt{2}$. $T \propto 1/\sqrt{g}$: pendulum clock runs slower at higher altitudes (smaller $g$). Second pendulum: $T = 2$ s, $L \approx 1$ m. Seconds pendulum is used to measure $g$: $g = 4\pi^2 L/T^2$.

4. Resonance in SHM

When driving frequency = natural frequency: resonance occurs. Amplitude becomes maximum (theoretically infinite without damping). Damping reduces amplitude at resonance. Quality factor $Q = \omega_0/(2\gamma)$ ($\gamma$ = damping coefficient). High $Q$: sharp resonance. Examples: pushing a swing at natural frequency. Breaking a wine glass with sound. Tacoma Narrows bridge collapse (wind resonance). Magnetic resonance imaging (MRI) uses nuclear spin resonance.

5. Energy in SHM

Total mechanical energy: $E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$ (constant, conserved). $KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2-x^2) = E\cos^2(\omega t)$. $PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2 = E\sin^2(\omega t)$. KE + PE = constant = $E$. At mean position: $KE = E$ (maximum), $PE = 0$. At extremes: $KE = 0$, $PE = E$ (maximum). Average KE = Average PE = $E/2$.

6. Damped Oscillations

With resistive force $F_{damp} = -bv$: $x = Ae^{-bt/(2m)}\cos(\omega't + \phi)$. Damped frequency: $\omega' = \sqrt{\omega_0^2 - (b/2m)^2}$. Three cases: Underdamped ($b < 2m\omega_0$): oscillates with decreasing amplitude. Critically damped ($b = 2m\omega_0$): returns to equilibrium fastest without oscillating (desired in car suspension, galvanometer). Overdamped ($b > 2m\omega_0$): very slowly returns to equilibrium. Car shock absorbers: near critical damping.

7. Superposition of SHMs

Same frequency, same direction: resultant amplitude $A = \sqrt{A_1^2+A_2^2+2A_1A_2\cos\delta}$ where $\delta$ = phase difference. In phase ($\delta=0$): $A = A_1+A_2$. Antiphase ($\delta=\pi$): $A = |A_1-A_2|$. Perpendicular SHMs of same frequency: Lissajous figures (circles, ellipses, lines depending on phase). Perpendicular SHMs of different frequencies: complex Lissajous patterns (used to compare frequencies).

8. Wave Motion and SHM

Progressive wave: each particle performs SHM with phase increasing with distance. $y = A\sin(\omega t - kx)$ for wave moving in $+x$. Wavelength $\lambda = 2\pi/k$. Speed $v = \omega/k = f\lambda$. Intensity $I \propto A^2$. Standing wave: $y = 2A\cos(kx)\sin(\omega t)$. Amplitude varies with position: nodes at $kx = n\pi$, antinodes at $kx = (n+1/2)\pi$. Standing waves in pipes and strings produce resonant frequencies discussed in wave theory.

Frequently Asked Questions
1. Why is maximum velocity proportional to amplitude?
At mean position ($x=0$), all energy is kinetic: $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$. So $v_{max} = A\sqrt{k/m} = A\omega$. This means: if amplitude doubles, maximum speed doubles (for same spring/mass). Amplitude $A$ and angular frequency $\omega$ together determine $v_{max}$. For different spring constants: same amplitude but stiffer spring ($k$ larger) gives higher $\omega$ and thus higher $v_{max}$.
2. Why does the period of a vertical spring equal that of a horizontal spring?
For vertical spring: at equilibrium, spring stretched by $\delta = mg/k$. If displaced by $x$ from equilibrium: net force $= -kx$ (the $mg$ and $k\delta$ cancel). So restoring force is still $-kx$, same as horizontal. Period $T = 2\pi\sqrt{m/k}$ is the same. Gravity only shifts the equilibrium position, not the oscillation period. This is why laboratory pendulum clocks and spring watches work at same rate regardless of orientation.
3. How does cutting a spring affect its spring constant?
Spring constant $k \propto 1/L$ (inversely proportional to length). If a spring of constant $k$ is cut into $n$ equal pieces: each piece has length $L/n$, so $k_{piece} = nk$. If these $n$ pieces are connected in parallel: $k_{eff} = n \times nk = n^2k$. If connected in series: $1/k_{eff} = n/(nk) = 1/k$, so $k_{eff} = k$ (same as original). This makes sense: cutting and reconnecting in series gives back original spring.
4. What is the condition for SHM?
A system executes SHM if: (1) The restoring force is proportional to displacement from equilibrium: $F = -kx$ (linear restoring force). (2) The restoring force is directed toward equilibrium. The general equation of SHM: $d^2x/dt^2 + \omega^2 x = 0$. Solution: $x = A\sin(\omega t + \phi)$. Many physical systems approximate SHM for small displacements: pendulum (for small angles), spring, vibrating string, LC circuit (electrical SHM), molecular vibrations.
5. What is the phase relationship between displacement, velocity and acceleration in SHM?
$x = A\sin(\omega t)$. $v = A\omega\cos(\omega t) = A\omega\sin(\omega t + 90°)$. Velocity leads displacement by 90°. $a = -A\omega^2\sin(\omega t) = A\omega^2\sin(\omega t + 180°)$. Acceleration leads displacement by 180° (i.e., acceleration is opposite to displacement). At positive extreme: $x = +A$, $v = 0$, $a = -A\omega^2$ (max, toward equilibrium). At mean position: $x = 0$, $v = v_{max}$, $a = 0$.
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