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.}