Questions & explanations
1. Which of the following displacement functions represents simple harmonic motion?
- x = sin ωt + sin 2ωt
- x = sin² ωt
- x = sin ωt + cos ωt
- x = sin ωt + cos 2ωt
Answer: x = sin ωt + cos ωt
Simple harmonic motion (SHM) is a periodic motion where displacement is a single sine or cosine function of time with a unique angular frequency. x = sin ωt + cos ωt can be written as √2 sin(ωt + π/4), which has a single ω. The other options contain multiple frequencies (ω and 2ω) or a squared term that reduces to a constant plus cos 2ωt, so they are periodic but not SHM.
2. Which of the following is an example of periodic motion that is NOT oscillatory?
- A rotating ceiling fan
- A vibrating tuning fork
- A pendulum swinging
- A bouncing ball on a spring
Answer: A rotating ceiling fan
Periodic motion repeats at regular intervals. Oscillatory motion is to-and-fro about a mean position. A rotating ceiling fan is periodic (each blade returns after a fixed time) but does not oscillate because it does not move to and fro about a mean position. The other options all involve to-and-fro motion about a mean position, so they are oscillatory.
3. A lightly damped oscillator has natural frequency 10 rad/s and is driven by a force of constant amplitude. If damping is increased, what happens to the resonance amplitude and width?
- Amplitude decreases, width decreases
- Amplitude increases, width decreases
- Amplitude decreases, width increases
- Amplitude increases, width increases
Answer: Amplitude decreases, width increases
For a driven damped oscillator, resonance amplitude A_max = F₀/(bω₀) is inversely proportional to damping coefficient b, so increasing damping decreases amplitude. The full width at half maximum (Δω) is proportional to b, so width increases. Hence amplitude decreases and width increases.
4. In the mechanical-electrical analogy, which pair is analogous to mass and spring constant?
- Inductance and capacitance
- Inverse inductance and capacitance
- Resistance and capacitance
- Inductance and inverse capacitance
Answer: Inductance and inverse capacitance
In the force-voltage analogy, comparing the differential equations: m d²x/dt² + kx = F for a spring-mass system and L d²q/dt² + (1/C)q = V for an LC circuit. Thus, mass m is analogous to inductance L, and spring constant k is analogous to inverse capacitance 1/C.
5. Which of the following systems executes simple harmonic motion?
- A particle in potential U = kx^4
- A pendulum with large amplitude
- A mass on a spring with F = -kx
- A ball bouncing elastically on a floor
Answer: A mass on a spring with F = -kx
For SHM, restoring force must be proportional to displacement and opposite. F = -kx satisfies this. Other options have nonlinear forces: pendulum large angle gives sinθ ≈ θ - θ^3/6, U = kx^4 gives F = -4kx^3, bouncing ball has constant force between impacts.
6. Two springs of force constants 100 N/m and 200 N/m are connected in series. A 1 kg mass is attached. What is the time period of oscillation?
- 0.77 s
- 0.36 s
- 0.54 s
- 0.89 s
Answer: 0.77 s
For springs in series, effective spring constant k_eff is given by 1/k_eff = 1/k1 + 1/k2 = 1/100 + 1/200 = 3/200, so k_eff = 200/3 ≈ 66.67 N/m. Time period T = 2π√(m/k_eff) = 2π√(1/(200/3)) = 2π√(3/200) = 2π√0.015 ≈ 2π×0.1225 ≈ 0.77 s.
7. Two springs of constants k₁ = 100 N/m and k₂ = 400 N/m are used with a mass of 1 kg. Which arrangement gives the smallest time period?
- Series
- Parallel
- Both parallel and both-sides give same smallest T
- Both-sides (block between two walls)
Answer: Both parallel and both-sides give same smallest T
For parallel: k_eff = k₁ + k₂ = 500 N/m, T = 2π√(1/500) ≈ 0.281 s. For both-sides: k_eff = k₁ + k₂ = 500 N/m, same T. For series: k_eff = (100×400)/(500) = 80 N/m, T ≈ 0.703 s. So parallel and both-sides give the same smallest T.
8. A 2 kg mass hangs from a vertical spring of force constant 200 N/m. What is the time period of vertical oscillations?
- 0.89 s
- 0.20 s
- 0.45 s
- 0.63 s
Answer: 0.63 s
For a vertical spring-mass system, gravity only shifts equilibrium. The restoring force is -kx, giving ω = √(k/m). Here k = 200 N/m, m = 2 kg, so ω = √(200/2) = 10 rad/s. Time period T = 2π/ω = 2π/10 ≈ 0.628 s ≈ 0.63 s.
9. A particle in SHM has amplitude 10 cm and angular frequency 2 rad/s. What is its speed when displacement is 6 cm?
- 16 cm/s
- 8 cm/s
- 12 cm/s
- 20 cm/s
Answer: 16 cm/s
Using conservation of mechanical energy in SHM: total energy = (1/2)mω²A². At displacement x, kinetic energy = total energy - potential energy = (1/2)mω²(A² - x²). Thus v² = ω²(A² - x²) = 4×(100-36)=256, so v=16 cm/s.
10. A particle of mass 0.5 kg moves in a potential U(x) = 2x² + x⁴ J (x in m). For small oscillations about equilibrium, what is its time period?
- π s
- π/√2 s
- 2π s
- π/2 s
Answer: π/√2 s
For small oscillations, approximate U(x) ≈ 2x² near equilibrium at x=0. Compare with U = ½ k x² to get k = 4 N/m. Mass m = 0.5 kg, so ω = √(k/m) = √(4/0.5) = √8 = 2√2 rad/s. Time period T = 2π/ω = 2π/(2√2) = π/√2 s.
11. In a damped oscillator, the amplitude decays to half its initial value in 2.0 s. If m = 0.5 kg, what is the damping constant b?
- 0.693 N·s/m
- 0.347 N·s/m
- 0.173 N·s/m
- 0.500 N·s/m
Answer: 0.347 N·s/m
For a damped oscillator, amplitude A(t) = A₀ e^{-b t/(2m)}. Given A(2)/A₀ = 1/2, so e^{-b·2/(2×0.5)} = e^{-2b} = 1/2. Taking natural log: -2b = ln(1/2) = -ln2, so b = (ln2)/2 ≈ 0.693/2 = 0.3465 N·s/m ≈ 0.347 N·s/m.
12. A 2 kg mass attached to a spring (k = 8 N/m) and a damper (b = 4 Ns/m) is driven by F = 10 cos(3t) N. What is the steady-state oscillation frequency of the mass?
- 2 rad/s
- 1 rad/s
- 4 rad/s
- 3 rad/s
Answer: 3 rad/s
In forced oscillations, the steady-state response oscillates at the driving frequency ω_d, not the natural frequency. Here ω_d = 3 rad/s from the forcing term 10 cos(3t). Hence the mass oscillates at 3 rad/s.