Simple harmonic motion, damping and resonance (RP7) – A-level Physics exam questions with mark scheme

1Core4 marks

A mass of 0.40 kg hangs from a spring of stiffness 25 N m$^{-1}$. The mass is pulled down 3.0 cm from its equilibrium position and released so that it oscillates with simple harmonic motion.

(a) Calculate the period of the oscillations. [2]

(b) Calculate the maximum speed of the mass. [2]

(a) $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.40}{25}}$(1)
(a) $T$ = 0.79 s (0.795)(1)
(b) $v_{max} = \omega A$ with $\omega = 2\pi/T$ = 7.9 rad s$^{-1}$ (or $\sqrt{k/m}$)(1)
(b) $v_{max} = 7.91 \times 0.030$ = 0.24 m s$^{-1}$ (ecf from (a))(1)
Answer: (a) 0.79 s (b) 0.24 m s$^{-1}$
Examiner tip: Amplitude must be in metres (0.030 m) before using v = ωA.

2Hard5 marks

The graph shows how the displacement $x$ of an oscillating object varies with time $t$. The object moves with simple harmonic motion.

(a) State one time shown on the graph at which the speed of the object is a maximum. [1]

(b) Calculate the maximum acceleration of the object. [2]

(a) any one of 0.20 s, 0.60 s, 1.0 s, 1.4 s (where $x = 0$)(1)
(b) period read as 0.80 s, so $\omega = 2\pi/0.80$ = 7.85 rad s$^{-1}$(1)
(b) $a_{max} = \omega^2 A = 7.85^2 \times 0.040$ = 2.5 m s$^{-2}$ (2.47) (A must be in m)(1)
(c) $x = A\cos\omega t = 0.040\cos(7.85 \times 0.10)$ (calculator in radians)(1)
(c) $x$ = 0.028 m or 2.8 cm(1)
Answer: (a) e.g. 0.20 s (b) 2.5 m s$^{-2}$ (c) 2.8 cm
Examiner tip: Use x = A cos ωt here because the graph starts at maximum displacement, and put your calculator in radians.

3Standard4 marks

(a) A student wants to make a simple pendulum with a period of 2.0 s. Calculate the length of string needed. [2]

(b) The pendulum is now driven by a periodic force whose frequency can be varied. Describe how the amplitude–frequency graph for the pendulum changes when the damping is increased. [2]

(a) $L = \frac{gT^2}{4\pi^2} = \frac{9.81 \times 2.0^2}{4\pi^2}$(1)
(a) $L$ = 0.99 m (0.994)(1)
(b) peak (maximum) amplitude is lower (accept: amplitude is smaller at all frequencies)(1)
(b) peak is broader / flatter (accept: peak occurs at a slightly lower frequency)(1)
Answer: (a) 0.99 m (b) lower and broader resonance peak
Examiner tip: For damping questions give two separate changes to the graph: the height of the peak and its width.

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4Hard5 marks

A trolley of mass 0.25 kg is attached between springs on a horizontal track. The combined stiffness of the springs is 40 N m$^{-1}$. The trolley is displaced by 5.0 cm and released, and it oscillates with simple harmonic motion. Friction is negligible.

(a) Calculate the total energy of the oscillating system. [1]

(b) Calculate the speed of the trolley when its displacement is 3.0 cm. [2]

(c) The trolley is replaced by one of mass 1.0 kg, again displaced by 5.0 cm and released. Explain what happens to the maximum speed. [2]

(a) $E = \frac{1}{2}kA^2 = \frac{1}{2} \times 40 \times 0.050^2$ = 0.050 J(1)
(b) $\omega = \sqrt{k/m} = \sqrt{40/0.25}$ = 12.65 rad s$^{-1}$ and $v = \omega\sqrt{A^2 - x^2}$ (or ½kA² − ½kx² = ½mv²)(1)
(b) $v = 12.65\sqrt{0.050^2 - 0.030^2}$ = 0.51 m s$^{-1}$ (allow 0.50 m s$^{-1}$ from $\omega$ = 12.6)(1)
(c) total energy $\frac{1}{2}kA^2$ is unchanged (same $k$ and $A$) / $\omega$ halves because $\omega = \sqrt{k/m}$ and $m$ is 4×(1)
(c) so maximum speed halves (from 0.63 to 0.32 m s$^{-1}$) since $\frac{1}{2}mv_{max}^2$ is constant / $v_{max} = \omega A$(1)
Answer: (a) 0.050 J (b) 0.51 m s$^{-1}$ (c) maximum speed halves
Examiner tip: In 'explain' questions about changing mass, quote the relationship and state the factor (halves) — 'decreases' alone usually loses the second mark.

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