Simple Harmonic Motion, Damping and Resonance: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A particle moves in SHM along the x-axis about x=0. Its speed is 4πcm/s when x=+3cm and 3πcm/s when x=+4cm. At t=0 the particle is at x=+3cm moving in the −x direction. What is the earliest time t>0 at which the particle is at x=+4cm moving in the +x direction? Exactly one option is correct.
A0.5s
B1.0s
C1.5s
D1.75s
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Answer: C
From v2=ω2(A2−x2): ω2(A2−9)=16π2 and ω2(A2−16)=9π2. Subtracting, 7ω2=7π2, so ω=πrad/s (T=2s) and A=5cm.
Write x=5cos(πt+ϕ0), v=−5πsin(πt+ϕ0). At t=0: cosϕ0=3/5, and v<0 needs sinϕ0=4/5, so ϕ0=tan−1(4/3).
Required state: cosθ=4/5 with sinθ<0 (so v>0): θ=2π−tan−1(3/4).
πt=2π−tan−1(3/4)−tan−1(4/3)=2π−π/2, so t=1.5s.
Trap: 0.5s is when the particle first reaches x=−4cm (direction/sign slip).
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A block of mass m hangs from the axle of a light, frictionless movable pulley P. A light inextensible string has one end fixed to the ceiling; it runs vertically down, passes under P, runs vertically up over a light frictionless pulley Q fixed to the ceiling, and then runs vertically down to the upper end of a light vertical spring of force constant k whose lower end is fixed to the floor. A second light vertical spring, also of force constant k, connects the bottom of the block to the floor. All string segments stay vertical and taut, and the block moves only vertically. What is the period of small vertical oscillations of the block? Exactly one option is correct.
A2π5km
B2π3km
Cπkm
D2π2km
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Answer: A
Let the block (and P) move down by y from equilibrium. Each of the two string segments supporting P lengthens by y, so a length 2y of string is drawn over Q and the top of the first spring rises by 2y: its extension increases by 2y and the string tension increases by ΔT=2ky.
P is supported by two segments, so the upward force on the block increases by 2ΔT=4ky.
The lower spring is compressed by an extra y and pushes up with an extra ky.
Net restoring force =5ky, so my¨=−5ky and T=2πm/(5k).
Traps: πm/k=2πm/(4k) omits the lower spring; 2πm/(3k) uses factor 2 instead of 4 for the pulley.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A uniform rod of mass M and length L is pivoted at one end about a fixed horizontal frictionless axis perpendicular to the rod and makes small oscillations in a vertical plane. A small bead of mass m (any value m>0; treat it as a point mass) is now fixed on the rod at distance y from the pivot. For which non-zero value of y is the period of small oscillations the same as without the bead, whatever the value of m? Exactly one option is correct.
A2L
B32L
C3L
D43L
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Answer: B
Without bead: T02=4π2Mg(L/2)I=4π2MgL/2ML2/3=4π23g2L, i.e. equivalent simple-pendulum length ℓ0=2L/3.
With bead: T2=4π2g(ML/2+my)ML2/3+my2. Setting T=T0:
3ML2+my2=32L(2ML+my)⇒my2=32Lmy⇒y=32L (independent of m).
Insight: a point mass placed at the centre of oscillation (distance equal to the equivalent length I/(Md)) is itself a simple pendulum of the same period, so adding it changes nothing.
Trap: L/3 is the radius of gyration about the pivot, not the equivalent length.
Difficulty: Very hard
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