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Date	May 2017	Marks available	2	Reference code	17M.2.HL.TZ1.7
Level	Higher level	Paper	Paper 2	Time zone	1
Command term	Describe	Question number	7	Adapted from	N/A

Question

A student is investigating a method to measure the mass of a wooden block by timing the period of its oscillations on a spring.

A 0.52 kg mass performs simple harmonic motion with a period of 0.86 s when attached to the spring. A wooden block attached to the same spring oscillates with a period of 0.74 s.

With the block stationary a longitudinal wave is made to travel through the original spring from left to right. The diagram shows the variation with distance x of the displacement y of the coils of the spring at an instant of time.

A point on the graph has been labelled that represents a point P on the spring.

Describe the conditions required for an object to perform simple harmonic motion (SHM).

[2]

Calculate the mass of the wooden block.

[2]

In carrying out the experiment the student displaced the block horizontally by 4.8 cm from the equilibrium position. Determine the total energy in the oscillation of the wooden block.

[3]

A second identical spring is placed in parallel and the experiment in (b) is repeated. Suggest how this change affects the fractional uncertainty in the mass of the block.

[3]

State the direction of motion of P on the spring.

[1]

e.i.

Explain whether P is at the centre of a compression or the centre of a rarefaction.

[2]

e.ii.

Markscheme

acceleration/restoring force is proportional to displacement
and in the opposite direction/directed towards equilibrium

ALTERNATIVE 1

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{m_{1}}{m_{2}}$ $\frac{{T_1^2}}{{T_2^2}} = \frac{{{m_1}}}{{{m_2}}}$

mass = 0.38 / 0.39 «kg»

ALTERNATIVE 2

«use of T $= 2 π \sqrt{\frac{m}{k}}$ $= 2\pi \sqrt {\frac{m}{k}}$ » k = 28 «Nm^–1»

«use of T $= 2 π \sqrt{\frac{m}{k}}$ $= 2\pi \sqrt {\frac{m}{k}}$ » m = 0.38 / 0.39 «kg»

Allow ECF from MP1.

ω = « $\frac{2 π}{0.74}$ $\frac{{2\pi }}{{0.74}}$ » = 8.5 «rads^–1»

total energy = $\frac{1}{2} \times 0.39 \times {8.5}^{2} \times (4.8 \times 10^{- 2})^{2}$ $\frac{1}{2} \times 0.39 \times {8.5^2} \times {(4.8 \times {10^{ - 2}})^2}$

= 0.032 «J»

Allow ECF from (b) and incorrect ω.

Allow answer using k from part (b).

spring constant/k/stiffness would increase
T would be smaller
fractional uncertainty in T would be greater, so fractional uncertainty of mass of block would be greater

left

e.i.

coils to the right of P move right and the coils to the left move left

hence P at centre of rarefaction

Do not allow a bald statement of rarefaction or answers that don’t include reference to the movement of coils.

Allow ECF from MP1 if the movement of the coils imply a compression.

e.ii.