Level 1 — RecognitionOscillations & Waves

Oscillations & Waves

20 minutes30 marksprintable — key stays hidden on paper

Level 1: Recognition Test

Time limit: 20 minutes Total marks: 30 Instructions: Answer all questions. For True/False, a correct justification is required for full marks.


Section A — Multiple Choice (1 mark each) [10 marks]

Q1. In simple harmonic motion, the restoring force is: (a) constant in magnitude (b) proportional to displacement and directed towards equilibrium (c) proportional to velocity (d) proportional to acceleration squared

Q2. The angular frequency ω\omega, period TT and frequency ff are related by: (a) ω=2πf=2πT\omega = 2\pi f = \dfrac{2\pi}{T} (b) ω=f2π\omega = \dfrac{f}{2\pi} (c) ω=T2π\omega = \dfrac{T}{2\pi} (d) ω=2πT\omega = 2\pi T

Q3. For a mass on a spring in SHM, the velocity is maximum when: (a) x=Ax = A (b) x=A/2x = A/2 (c) x=0x = 0 (d) x=Ax = -A

Q4. The period of a simple pendulum of length LL is: (a) 2πg/L2\pi\sqrt{g/L} (b) 2πL/g2\pi\sqrt{L/g} (c) 12πL/g\dfrac{1}{2\pi}\sqrt{L/g} (d) 2πLg2\pi L g

Q5. In a longitudinal wave, the particle oscillations are: (a) perpendicular to wave propagation (b) parallel to wave propagation (c) circular (d) zero

Q6. The condition for constructive interference (path difference Δ\Delta) is: (a) Δ=(n+12)λ\Delta = (n+\tfrac{1}{2})\lambda (b) Δ=nλ\Delta = n\lambda (c) Δ=λ/4\Delta = \lambda/4 (d) Δ=nλ/3\Delta = n\lambda/3

Q7. A critically damped oscillator: (a) oscillates forever (b) returns to equilibrium in the shortest time without oscillating (c) overshoots many times (d) has no damping

Q8. The beat frequency produced by two waves of frequencies f1f_1 and f2f_2 is: (a) f1+f2f_1 + f_2 (b) f1f2\sqrt{f_1 f_2} (c) f1f2|f_1 - f_2| (d) 12(f1+f2)\tfrac{1}{2}(f_1+f_2)

Q9. The Mach number is defined as: (a) M=c/vM = c/v (sound speed / source speed) (b) M=v/cM = v/c (source speed / sound speed) (c) M=vcM = v \cdot c (d) M=vcM = v - c

Q10. A quantity that stays constant throughout the motion of an ideal SHM oscillator is: (a) kinetic energy (b) potential energy (c) total mechanical energy (d) displacement


Section B — Matching (1 mark each) [8 marks]

Q11. Match each term with its correct expression/description:

# Term Expression / Description
i SHM displacement P 12kA2\tfrac{1}{2}kA^2
ii Total energy in SHM Q x=Acos(ωt+ϕ)x = A\cos(\omega t + \phi)
iii Wave speed R point of zero displacement in a standing wave
iv Node S v=fλv = f\lambda

Q12. Match the oscillator/wave concept with its keyword:

# Concept Keyword
i Resonance P energy loss, amplitude decays
ii Damping Q driving frequency = natural frequency
iii Overtone R frequency above the fundamental
iv Q factor S sharpness of resonance / quality

Section C — True / False with Justification (2 marks each) [12 marks]

(1 mark correct T/F, 1 mark justification)

Q13. In SHM, acceleration is maximum at the equilibrium position.

Q14. For a transverse wave on a string, wave speed depends on the tension and the linear mass density.

Q15. At resonance in a lightly damped forced oscillator, the amplitude is at its maximum.

Q16. The frequency of a wave changes when it passes from one medium to another.

Q17. When a source of sound moves towards a stationary observer, the observed frequency is lower than the emitted frequency.

Q18. Doubling the amplitude of an SHM oscillator doubles its total energy.

Answer keyMark scheme & solutions

Section A — MCQ (1 mark each)

Q1 — (b). SHM is defined by F=kxF=-kx: force proportional to displacement, opposite in direction. (1)

Q2 — (a). ω=2πf\omega=2\pi f and f=1/Tf=1/T, so ω=2π/T\omega=2\pi/T. (1)

Q3 — (c). From v=ωA2x2v=\omega\sqrt{A^2-x^2}, vv is maximum (=ωA=\omega A) at x=0x=0. (1)

Q4 — (b). Small-angle pendulum: T=2πL/gT=2\pi\sqrt{L/g}. (1)

Q5 — (b). Longitudinal waves oscillate parallel to propagation (e.g. sound). (1)

Q6 — (b). Constructive interference: whole-number of wavelengths, Δ=nλ\Delta=n\lambda. (1)

Q7 — (b). Critical damping returns to equilibrium fastest without oscillating. (1)

Q8 — (c). Beat frequency =f1f2=|f_1-f_2|. (1)

Q9 — (b). M=v/cM=v/c = source speed ÷ sound speed. (1)

Q10 — (c). KE and PE interchange, but total E=12kA2E=\tfrac{1}{2}kA^2 is constant. (1)

Section B — Matching (1 mark each pairing)

Q11: i→Q, ii→P, iii→S, iv→R (1 each = 4)

Q12: i→Q, ii→P, iii→R, iv→S (1 each = 4)

Section C — True/False with Justification

Q13 — FALSE. (1) Acceleration a=ω2xa=-\omega^2 x is proportional to xx; it is zero at equilibrium and maximum at x=±Ax=\pm A. (1)

Q14 — TRUE. (1) Wave speed on a string v=T/μv=\sqrt{T/\mu}, depending on tension TT and linear density μ\mu. (1)

Q15 — TRUE. (1) Forced oscillation amplitude peaks when the driving frequency approaches the natural frequency (resonance), especially sharp for low damping (high Q). (1)

Q16 — FALSE. (1) Frequency is set by the source and is unchanged; wave speed and wavelength change between media (v=fλv=f\lambda, ff constant). (1)

Q17 — FALSE. (1) Source approaching → wavefronts compressed → observed frequency higher (f=fc/(cvs)f'=f\,c/(c-v_s)). (1)

Q18 — FALSE. (1) E=12kA2A2E=\tfrac{1}{2}kA^2\propto A^2, so doubling AA quadruples the energy. (1)

[
  {"claim":"v_max occurs at x=0 for v=omega*sqrt(A^2-x^2)",
   "code":"A,w,x=symbols('A w x',positive=True); v=w*sqrt(A**2-x**2); result = (v.subs(x,0)==w*A) and (v.subs(x,A)==0)"},
  {"claim":"Total SHM energy is constant = kA^2/2",
   "code":"k,A,w,t,phi=symbols('k A w t phi',positive=True); m=k/w**2; x=A*cos(w*t+phi); v=diff(x,t); E=simplify(Rational(1,2)*m*v**2+Rational(1,2)*k*x**2); result = simplify(E-Rational(1,2)*k*A**2)==0"},
  {"claim":"Doubling amplitude quadruples energy",
   "code":"k,A=symbols('k A',positive=True); E1=Rational(1,2)*k*A**2; E2=Rational(1,2)*k*(2*A)**2; result = simplify(E2/E1)==4"},
  {"claim":"Approaching source raises observed frequency",
   "code":"f,c,vs=symbols('f c vs',positive=True); fp=f*c/(c-vs); result = simplify(fp - f) .subs({f:1,c:340,vs:30}) > 0"}
]