A metal block hangs vertically from one end of a spring. The other end of the spring is tied to a thread that passes over a pulley and is attached to a vibrator, as shown in Fig. 4.1.

Question 3

A metal block hangs vertically from one end of a spring. The other end of the spring is tied to a thread that passes over a pulley and is attached to a vibrator, as shown in Fig. 4.1.

Fig. 4.1

(a) The vibrator is switched off.

The metal block of mass 120 g is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 4.2.

Fig. 4.2

For the vibrations of the block, calculate

(i) the angular frequency ω, [2]

(ii) the energy of the vibrations. [2]

(b) The vibrator is now switched on.

The frequency of vibration is varied from 0.7f to 1.3f where f is the frequency of vibration of the block in (a).

For the block, complete Fig. 4.3 to show the variation with frequency of the amplitude of

vibration. Label this line A. [3]

Fig. 4.3

(c) Some light feathers are now attached to the block in (b) to increase air resistance.

The frequency of vibration is once again varied from 0.7f to 1.3f. The new amplitude of

vibration is measured for each frequency.

On Fig. 4.3, draw a line to show the variation with frequency of the amplitude of vibration.

Label this line B. [2]

[Total: 9]

Reference: Past Exam Paper – June 2016 Paper 42 Q4

Solution:

(a)

(i)

{From the graph, period T = 0.6 s}

T = 0.60 s        and     ω = 2π / T

Angular frequency ω = 10 (10.47) rad s^-1

(ii)

{The energy of an oscillator can be obtained by}

energy = ½ m ω² x0²                    or  ½ mv2  and v = ωx0

{Quantities need to be in SI units.}

energy = ½ ×120×10^-3 × (10.5)² × (2.0×10^-2)²

energy = 2.6 × 10^-3 J                                               

(b)

sketch: smooth curve in correct directions   

peak at f                                                         

amplitude never zero and line extends from 0.7f to 1.3f

{When the frequency of the vibrator is equal to the frequency of vibration of the block (= f), resonance occurs and the amplitude of oscillation is greatest.}

(c)

sketch: peaked line always below a peaked line A                                       

peak not as sharp and at (or slightly less than) frequency of peak in line A

{The air resistance causes damping on the oscillation. This reduces the amplitude and causes resonance to occur at a slightly lower frequency with the peak being a bit flatter.}