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Q4. A seismometer is a device that is used to record the movement of the ground during an earthquake.

A simple seismometer is shown in the diagram below:

A heavy spherical ball is attached to a pivot by a rod so that the rod and ball can move in a vertical plane.

The rod is suspended by a spring so that, in equilibrium, the spring is vertical and the rod is horizontal.

A pen is attached to the ball.

The pen draws a line on graph paper attached to a drum rotating about a vertical axis.

Bolts secure the seismometer to the ground so that the frame of the seismometer moves during the earthquake.

(a) The ball is made of steel of density 8030 kg m–3 and has a diameter of 5.0 cm.

Show that the weight of the ball is approximately 5 N.

d = 5.0 cm = 5.0 x 10-2 m

r = ½ d = 2.5 x 10-2 m

Volume of a sphere = 4/3 π r3

V = 4/3 π r3

V = 4/3 π (2.5 x 10-2)3

V = 6.54 x 10-5 m3

 

density = mass/volume

ρ = m/V

m = ρV

m = 8030 x 6.54 x 10-5

m = 0.53 kg

 

weight = mass x gravitational field strength

w = mg

w = 0.53 x 9.81

Using g = 10 N/kg loses you a mark - use the A Level value from your data sheet!

w = 5.2 N

w ≈ 5N Q.E.D.

 

[3 marks]

(b) The distance from the surface of the ball to the pivot is 12.0 cm, as shown in the diagram.

Calculate the moment of the weight of the ball about the pivot when the rod is horizontal.

The centre of gravity/mass of the ball is at its centre. Therefore the distance for the moment to the pivot point is 12.0 cm + 2.5 cm = 14.5 cm = 0.145 m

moment = force x perpendicular distance from the pivot point

anticlockwise moment = 5.2 N x 0.145 m = 0.75 Nm

 

[3 marks]

(c) The spring is attached at a distance of 8.0 cm from the pivot, and the spring has a stiffness of 100 N m–1.

Calculate the extension of the spring when the rod is horizontal and the spring is vertical.

You may assume the mass of the pen and the mass of the rod are negligible.

To find the force acting on the spring

Taking moments about the pivot:

clockwise moment from spring = anticlockwise moment from ball

F x 0.080 = 0.75

F = 9.4 N

To find the spring extension:

F = k x

x = F/k

x = 9.4/100

x = 0.094 m

[3 marks]

(d) Before an earthquake occurs, the line being drawn on the graph paper is horizontal.

Explain what happens to the line on the graph paper when an earthquake is detected and the frame of the seismometer accelerates rapidly downwards.

The line/pen moves up.

The examiner was told to ignore subsequent motion details - you are only asked what happens when the frame of the seismometer accelerates rapidly downwards. ‘Diagonally up’ and ‘up then down’ got you the mark but ‘up and down’ was rejected.

The ball is heavy and therefore has a lot of inertia. Therefore the downwards acceleration of the ball is much less than that of the frame and the ball does not move very far in the time taken for the frame (and the recotating drum) to move down.

[2 marks]

(Total 11 marks)