A spring is hung vertically from a fixed point. A mass M is hung from the other end of the spring, as illustrated in Fig. 3.1.

Question 4

A spring is hung vertically from a fixed point. A mass M is hung from the other end of the spring, as illustrated in Fig. 3.1.

Fig. 3.1

The mass is displaced downwards and then released. The subsequent motion of the mass is simple harmonic.

The variation with time t of the length L of the spring is shown in Fig. 3.2.

Fig. 3.2

(a) State:

(i) one time at which the mass is moving with maximum speed [1]

(ii) one time at which the spring has maximum elastic potential energy. [1]

(b) Use data from Fig. 3.2 to determine, for the motion of the mass:

(i) the angular frequency ω [2]

(ii) the maximum speed [2]

(iii) the magnitude of the maximum acceleration. [2]

(c) The mass M is now suspended from two springs, each identical to that in Fig. 3.1, as shown in Fig. 3.3.

Fig. 3.3

Suggest and explain the change, if any, in the period of oscillation of the mass. A numerical answer is not required. [2]

 [Total: 10]

Reference: Past Exam Paper – June 2019 Paper 42 Q3

Solution:

(a)

(i)

{In s.h.m., the mass moves with maximum speed at the equilibrium position – here, this corresponds to any time where L = 12 cm}

0.10 s              or 0.30 s                      or 0.50 s          or 0.70 s          or 0.90 s

(ii)

{Maximum elastic potential energy occurs when the spring is at its longest – this corresponds to the lowest position.}

0          or 0.40 s          or 0.80 s

{At the shortest length, the spring is compressed and in this situation, the mass also has GPE.}

(b)

(i)

ω = 2π / T

ω = 2π / 0.40

ω = 16 rad s^-1 

(ii)

{The maximum displacement x0 (amplitude) is 2.5 cm.}

v0 = ωx0

v0 = 15.7 × 2.5 × 10^-2

v0 = 0.39 m s^-1

(iii)

{Acceleration a = ω²x

The acceleration a is greatest when displacement x is maximum (that is, amplitude)}

a0 = ω²x0

a0 = (15.7² × 2.5 × 10^-2)

a0 = 6.2 m s^-2

or

a0 = ωv0             

a0 = 15.7 × 0.39

a0 = 6.2 m s^-2  

(c) The period of oscillation decreases as the acceleration is greater (for any given extension).

{A parallel combination of springs has a greater spring constant k.

From Hooke’s law: F = ke

With a greater spring constant, the restoring force F is greater (for any given extension).

With the 2 springs, when the mass M is positioned (without oscillating), the springs are extended. But this extension is smaller than before as the spring constant is now greater. This position is the equilibrium position. So, using 2 springs in parallel results in a new equilibrium position of the mass.

The amplitude of the oscillations is unconnected to the number of springs used.

To compare with the case of a single spring, for any given amplitude (similar in both case), the mass would have greater energy with 2 springs (elastic energy = ½ kx² and k is now greater while x is taken to be the same in both case). As the mass moves towards the equilibrium position, this energy is converted into KE. So, a greater amount of elastic potential energy means that more energy is converted into KE, and so, the mass would have a greater maximum speed.}

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.}