Oscillations and Waves — Easy Practice (Set 1)

Quick Recap — Simple Harmonic Motion

• SHM: the acceleration $a = -\omega^2 x$ is directed toward the mean position; $x = A\sin(\omega t + \phi)$.
• Frequency and period: $\omega = 2\pi f = \dfrac{2\pi}{T}$.
• Speeds: $v = \omega\sqrt{A^2 - x^2}$, with $v_{max} = A\omega$ and $a_{max} = A\omega^2$.
• Energy: total $E = \tfrac12 kA^2 = \tfrac12 m\omega^2 A^2$ (constant); KE is maximum at the mean position, PE at the extremes.
• Standard systems: mass-spring $T = 2\pi\sqrt{m/k}$; simple pendulum $T = 2\pi\sqrt{L/g}$.

Beyond-NCERT JEE Formulae

Basic SHM kinematics ($x=A\sin(\omega t+\phi)$, $v=\omega\sqrt{A^2-x^2}$, $a=-\omega^2 x$) are assumed known. This sheet collects the higher-yield results that JEE Main tests beyond the NCERT core.

1. Spring combinations and two-body SHM

• Series (springs end to end, same force, extensions add): $\dfrac{1}{k_{eff}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$, so $k_{eff}=\dfrac{k_1 k_2}{k_1+k_2}$ (softer than either spring).
• Parallel (springs side by side, same extension, forces add): $k_{eff}=k_1+k_2$ (stiffer than either spring).
• Spring cut to a fraction $f$ of its length: the constant rises to $\dfrac{k}{f}$; cutting a spring in half gives $2k$.
• Two-body SHM (two masses joined by one spring, free on a smooth surface): oscillates with the reduced mass $\mu=\dfrac{m_1 m_2}{m_1+m_2}$, so $T=2\pi\sqrt{\dfrac{\mu}{k}}$. The blocks swing about a fixed centre of mass with amplitudes in the inverse ratio $\dfrac{A_1}{A_2}=\dfrac{m_2}{m_1}$.

[JEE Tip] Read the geometry, not the picture. "End to end" is series; "side by side" is parallel. A block held between two walls by two springs is parallel ($k_{eff}=k_1+k_2$) even though the springs look opposed, because any displacement stretches one and compresses the other and both push back together.

2. Pendulums beyond the simple pendulum

• Physical (compound) pendulum (any rigid body swinging about a pivot): $T=2\pi\sqrt{\dfrac{I}{mgd}}$, where $I$ is the moment of inertia about the pivot and $d$ is the pivot-to-centre-of-mass distance. The equivalent simple-pendulum length is $L_{eq}=\dfrac{I}{md}$.
• Torsional pendulum (a disc or rod twisting on a wire): $T=2\pi\sqrt{\dfrac{I}{C}}$, with $C$ the torsional constant (restoring torque per radian).
• Effective gravity: in a lift accelerating up or down use $g_{eff}=g\pm a$; with a horizontal force giving acceleration $a$ (a charged bob in a field), the string hangs at $\tan\theta=\dfrac{a}{g}$ and $g_{eff}=\sqrt{g^2+a^2}$. In every case $T=2\pi\sqrt{\dfrac{L}{g_{eff}}}$.

[JEE Tip] For a uniform rod pivoted at one end, $I=\dfrac{mL^2}{3}$ and $d=\dfrac{L}{2}$, so $T=2\pi\sqrt{\dfrac{2L}{3g}}$ and $L_{eq}=\dfrac{2L}{3}$ — the mass cancels, exactly as for a simple pendulum.

3. Energy, timing and combining SHMs

• Energy split: total $E=\tfrac12 m\omega^2 A^2=\tfrac12 kA^2$ is constant, with $KE=\tfrac12 m\omega^2(A^2-x^2)$ and $PE=\tfrac12 m\omega^2 x^2$. Hence $KE=PE$ at $x=\dfrac{A}{\sqrt2}$, while at $x=\dfrac{A}{2}$ the split is $KE:PE=3:1$.
• Timing (started from the mean position, $x=A\sin\omega t$): mean to $\dfrac{A}{2}$ takes $\dfrac{T}{12}$; mean to $\dfrac{A}{\sqrt2}$ takes $\dfrac{T}{8}$; mean to $A$ takes $\dfrac{T}{4}$.
• From two speed-displacement readings: $\omega=\sqrt{\dfrac{v_1^2-v_2^2}{x_2^2-x_1^2}}$, and combining with $v_{max}$ and $a_{max}$ gives $\omega=\dfrac{a_{max}}{v_{max}}$, $A=\dfrac{v_{max}^2}{a_{max}}$.
• Superposition of two collinear SHMs of equal $\omega$ and phase gap $\phi$: resultant amplitude $A=\sqrt{A_1^2+A_2^2+2A_1 A_2\cos\phi}$.

[JEE Tip] SHM energy scales as $A^2$ and as $\omega^2$, so doubling the frequency at fixed amplitude quadruples the energy. When a problem gives both a speed and an acceleration at known points, form ratios to cancel the unknown $\omega$ before solving.

4. Damped and forced oscillations

• Light damping (drag $-bv$): amplitude decays as $A(t)=A_0 e^{-bt/2m}$; since $E\propto A^2$, energy decays as $E(t)=E_0 e^{-bt/m}$, twice as fast in the exponent.
• Quality factor: $Q=\dfrac{m\omega_0}{b}$; a larger $Q$ means slower decay and a sharper resonance peak.
• Resonance: a driven oscillator responds most strongly when the driving frequency nears the natural value $\omega_0=\sqrt{\dfrac{k}{m}}$.

[JEE Tip] "Amplitude halves" and "energy halves" happen at different times: $t_{A/2}=\dfrac{2m}{b}\ln 2$ but $t_{E/2}=\dfrac{m}{b}\ln 2$. Always check which quantity the question is halving.

5. Waves on strings and wave power

• Transverse wave speed on a stretched string: $v=\sqrt{\dfrac{T}{\mu}}$, where the linear mass density is $\mu=\dfrac{m}{L}$ (mass per unit length).
• String fixed at both ends (nodes at both ends, all harmonics present): $f_n=\dfrac{nv}{2L}=\dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$ for $n=1,2,3,\ldots$
• Power carried by a travelling wave: $P=\tfrac12\mu\omega^2 A^2 v$, so intensity scales as $\omega^2 A^2$.
• Junction reflection: entering a denser (slower) string the reflected pulse inverts ($v_2<v_1$); a free end reflects with no inversion.

[JEE Tip] Because $f\propto\sqrt{T}$ on a sonometer, a small tension change produces a doubled fractional frequency change: $\dfrac{\Delta f}{f}\approx\tfrac12\dfrac{\Delta T}{T}$. This is the shortcut behind sonometer-beat problems.

6. Organ pipes, end correction and beats

• Open pipe (antinodes at both ends, all harmonics): $f_n=\dfrac{nv}{2L}$ for $n=1,2,3,\ldots$
• Closed pipe (node at the closed end, antinode at the open end, odd harmonics only): $f_n=\dfrac{(2n-1)v}{4L}$, giving the sequence $f_1,3f_1,5f_1,\ldots$
• End correction: the open-end antinode sits just beyond the pipe, so replace $L$ by $L+e$ with $e\approx 0.6r$ per open end ($r$ = pipe radius). A resonance-tube pair of lengths cancels $e$ via $v=2f(\ell_2-\ell_1)$.
• Beats: two close frequencies give $f_{beat}=|f_1-f_2|$; loading a fork with wax lowers its frequency, which resolves the sign ambiguity.

[JEE Tip] A closed pipe of length $L$ has the same fundamental as an open pipe of length $2L$. For a closed pipe, successive resonances differ by $2f_1$, not $f_1$ — use that spacing to recover the fundamental from two observed resonance frequencies.

7. Doppler effect

• General form (medium at rest): $f'=f\dfrac{v\pm v_o}{v\mp v_s}$. Fix the signs so that approach raises pitch: use the upper signs when the observer moves toward the source and the source moves toward the observer.
• Only the source moves: $f'=f\dfrac{v}{v\mp v_s}$. Only the observer moves: $f'=f\dfrac{v\pm v_o}{v}$.
• Reflection off a moving wall: treat the wall first as a moving observer, then as a moving source re-emitting. An approaching wall of speed $u$ returns $f'=f\dfrac{v+u}{v-u}$, which beats with the original note at $f'-f$.

[JEE Tip] Equal source speed and observer speed do not give equal shifts: a source approaching at $u$ raises the pitch more than an observer approaching at $u$, because the source speed sits in the denominator. Only when $u\ll v$ do the two nearly agree.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Springs in series

Q. Two springs of force constants $k_1=400$ N/m and $k_2=600$ N/m are joined end to end, and a block of mass $2.4$ kg hangs from the free end. Find the period of small oscillations (take $\pi=3.14$).

Solution. End to end means the same force acts through both while their extensions add, so the springs combine in series: $$\frac{1}{k_{eff}}=\frac{1}{400}+\frac{1}{600}=\frac{3+2}{1200}=\frac{5}{1200},$$ which gives $k_{eff}=240$ N/m. The period is $$T=2\pi\sqrt{\frac{m}{k_{eff}}}=2\pi\sqrt{\frac{2.4}{240}}=2\pi\sqrt{0.01}=2\pi(0.1).$$ So $T=0.628$ s. A series pack is softer than either spring, giving a longer period.

Example 2 — Springs in parallel

Q. The same two springs ($k_1=400$ N/m, $k_2=600$ N/m) now act side by side on a block of mass $2.5$ kg on a smooth floor. Find the period ($\pi=3.14$).

Solution. Side by side means equal extension with forces adding, so the springs are in parallel: $$k_{eff}=k_1+k_2=400+600=1000,$$ i.e. $k_{eff}=1000$ N/m. Then $$T=2\pi\sqrt{\frac{2.5}{1000}}=2\pi\sqrt{0.0025}=2\pi(0.05).$$ So $T=0.314$ s. The parallel pack is stiffer than the series pack of Example 1, so it oscillates about twice as fast.

Example 3 — Two-block system and reduced mass

Q. Blocks of mass $m_1=1$ kg and $m_2=3$ kg on a smooth floor are joined by a spring of constant $k=300$ N/m. They are pulled apart and released. Find the period and the ratio of their amplitudes ($\pi=3.14$).

Solution. A spring free at both ends oscillates with the reduced mass: $$\mu=\frac{m_1 m_2}{m_1+m_2}=\frac{1\times 3}{1+3}=\frac{3}{4}=0.75,$$ so $\mu=0.75$ kg. The angular frequency and period are $$\omega=\sqrt{\frac{k}{\mu}}=\sqrt{\frac{300}{0.75}}=\sqrt{400}=20,\qquad T=\frac{2\pi}{\omega}=\frac{6.28}{20}.$$ So $\omega=20$ rad/s and $T=0.314$ s. Momentum stays zero about the fixed centre of mass, so $\dfrac{A_1}{A_2}=\dfrac{m_2}{m_1}=\dfrac{3}{1}$: the lighter block swings with three times the amplitude of the heavier one.

Example 4 — Physical (compound) pendulum

Q. A uniform rod of length $0.6$ m is pivoted at one end and swings in a vertical plane. Taking $g=10$ m/s$^2$ and $\pi=3.14$, find its period and the length of the simple pendulum that keeps the same time.

Solution. For a rod about one end, $I=\dfrac{mL^2}{3}$ and the centre of mass is at $d=\dfrac{L}{2}$. The physical-pendulum period is $$T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{mL^2/3}{mg(L/2)}}=2\pi\sqrt{\frac{2L}{3g}}.$$ Substituting the numbers, $$T=2\pi\sqrt{\frac{2(0.6)}{3(10)}}=2\pi\sqrt{\frac{1.2}{30}}=2\pi\sqrt{0.04}=2\pi(0.2).$$ So $T=1.256$ s. The equivalent simple-pendulum length is $L_{eq}=\dfrac{2L}{3}=0.4$ m, and indeed $2\pi\sqrt{0.4/10}=1.256$ s. The rod's mass cancels throughout.

Example 5 — Beats and an unknown fork

Q. A tuning fork $A$ of frequency $320$ Hz gives $4$ beats per second with a fork $B$. When a little wax is stuck onto $B$, the beat rate falls to $2$ per second. Find the original frequency of $B$.

Solution. The beat rate fixes only the size of the gap: $$f_{beat}=|f_A-f_B|=4,$$ so $f_B=320\pm 4$, that is $316$ Hz or $324$ Hz. Loading a fork with wax lowers its frequency. If $f_B$ were $316$ Hz, lowering it further would move it away from $320$ Hz and the beats would rise. Since the beats instead fall (from $4$ to $2$), $f_B$ must lie above $A$, so the original frequency is $f_B=324$ Hz. As a check, wax that drops $B$ from $324$ Hz to about $322$ Hz gives $|320-322|=2$ beats per second.

Example 6 — Open and closed organ pipes

Q. An open pipe and a closed pipe each have length $0.85$ m, and the speed of sound is $340$ m/s. Find the fundamental and the first two overtones of each pipe.

Solution. For the open pipe every harmonic is present, $f_n=\dfrac{nv}{2L}$: $$f_1=\frac{v}{2L}=\frac{340}{2(0.85)}=\frac{340}{1.7}=200.$$ So the fundamental is $200$ Hz, and the first two overtones are $2f_1=400$ Hz and $3f_1=600$ Hz. For the closed pipe only odd harmonics survive, $f_n=\dfrac{(2n-1)v}{4L}$: $$f_1=\frac{v}{4L}=\frac{340}{4(0.85)}=\frac{340}{3.4}=100.$$ So its fundamental is $100$ Hz, and the surviving overtones are $3f_1=300$ Hz and $5f_1=500$ Hz. The closed pipe sounds an octave below the open pipe of equal length and skips every even harmonic.

Example 7 — Doppler effect, moving source

Q. A siren of frequency $500$ Hz moves in a straight line at $30$ m/s. The speed of sound is $330$ m/s. Find the frequency heard by a stationary listener (a) as the siren approaches and (b) after it has passed.

Solution. Only the source moves, so $f'=f\dfrac{v}{v\mp v_s}$ (lower sign, $v-v_s$, for approach).

(a) Approaching: $$f'=500\times\frac{330}{330-30}=500\times\frac{330}{300}=500\times 1.1.$$ So $f'=550$ Hz.

(b) Receding: $$f'=500\times\frac{330}{330+30}=500\times\frac{330}{360}=500\times 0.9167.$$ So $f'=458.3$ Hz. The pitch jumps from $550$ Hz to $458.3$ Hz as the siren passes.

Example 8 — Doppler effect, moving observer

Q. Now the $500$ Hz source is stationary and the listener moves at $30$ m/s along the line joining them, with sound speed $330$ m/s. Find the frequency heard (a) approaching and (b) receding, and compare with Example 7.

Solution. Only the observer moves, so $f'=f\dfrac{v\pm v_o}{v}$ (upper sign, $v+v_o$, for approach).

(a) Approaching: $$f'=500\times\frac{330+30}{330}=500\times\frac{360}{330}=500\times 1.0909.$$ So $f'=545.5$ Hz.

(b) Receding: $$f'=500\times\frac{330-30}{330}=500\times\frac{300}{330}=500\times 0.9091.$$ So $f'=454.5$ Hz.

Compare with Example 7: at the same speed of $30$ m/s, a moving source gave $550$ Hz on approach but a moving observer gives only $545.5$ Hz. The two formulas are not interchangeable — the source speed sits in the denominator, so it shifts the pitch a little more.