Q1

(a) A body is moving with simple harmonic motion. State two conditions that must be satisfied concerning the acceleration of the body.

For a body to be moving with simple harmonic motion acceleration must be:

• proportional to displacement
• in the opposite direction to displacement, OR always acting towards a fixed point, or towards the centre of oscillation

(2 marks)

(b) A mass is suspended from a vertical spring and the system is allowed to come to rest. When the mass is now pulled down a distance of 76 mm and released, the time taken for 25 oscillations is 23 s.

Calculate

(i) the frequency of the oscillations,

25 oscillations in 23s - therefore time for one is

23/25 (that would be the period of the oscillation.

f = 1/T so

frequency = 25/23 = 1.09 (this is needed for any carry forward calculations)

f = 1.1 Hz (final answer should be quoted to 2sf)

(ii) the maximum acceleration of the mass,

a = (2πf)²A

a = (2π × 1.09)² × 76 × 10⁻³

a = 3.6 m s⁻²

(iii) the displacement of the mass from its rest position 0.60 s after being released. State the direction of this displacement.

x = A cos (2πft)

x = 76 × 10⁻³ cos (2π × 1.09 × 0.60)

x = − 4.3 × 10⁻² m

x = 43 mm

(use of f = 1.1 Hz gives x = (−)4.07 × 10⁻² m which rounds to 41 mm)

direction: above equilibrium position or upwards

(6 marks)

(c)The diagram below shows qualitatively how the velocity of the mass varies with time over the first two cycles after release.

(i) Using the axes below, sketch a graph to show qualitatively (this means they don't expect you to put numbers on the axes) how the displacement of the mass varies with time during the same time interval.

correct shape, i.e. cos curve

correct phase i.e. −(cos)

(ii) Using the axes in below sketch a graph to show qualitatively how the potential energy of the mass-spring system varies with time during the same time interval.

graph to show:

two cycles per oscillation

correct shape (even if phase is wrong) the shape is that of the sine squared function. Remember that kinetic energy and potential energy interchange - KE = ¹/₂ mv², so the shape of the energy graph (with time) is that of v².

correct starting point (i.e. full amplitude)

You cannot have 'negative energy' therefore the graph should not go into the negative part of the axes.

The potential energy will be at a maximum each time the spring is fully stretched and fully compressed. It will be zero when the spring is at equilibrium position. It starts at the fully stretched so should start at a maximum and should be zero each time displacement is zero (midpoint of the markings on the axes they gave you)...

(4 marks)

(Total 12 marks)