## Thursday, March 11, 2010

### Questions (MCQ) on Oscillations of a Spring-mass System

The following questions involving a spring-mass system are meant for checking whether you have a thorough understanding of the simple harmonic oscillations in which the restoring force is supplied by a spring:

(1) The frequency of vertical oscillations of a mass suspended at the end of a light spring is n. If the system is taken to a location where the acceleration due to gravity is reduced by 0.1%, the frequency of oscillation will be

(a) 1.01 n

(b) 0.99 n

(c) 1.001 n

(d) 0.999 n

(e) n

A spring-mass system (unlike the simple pendulum) does not require a gravitational force for oscillations since the restoring force required for oscillations is supplied by the elastic forces in the spring.

[Note that the period (T) of oscillations is given by T = 2π√(m/k) where m is the mass attached to the spring and k is the spring constant].

Therefore the change in ‘g’ does not affect the frequency and the correct option is (e).

(2) One end of a light spring is fixed to the ceiling and a mass M is suspended at the other end. When an additional mass m is attached to the mass M, the additional extension in the spring is e. The period of vertical oscillation of the spring-mass system now is

(a) 2π√[(M+m)e/mg]

(b) 2π√(me/mg)

(c) 2π√[(M+m) /mge]

(d) 2π√[me/(M+m)g]

(e) 2π√(M/mge)

The period (T) of oscillations is given by T = 2π√[(M+m)/k] where k is the spring constant.

Since an additional weight mg attached to the spring produces an additional extension e, the spring constant k = mg/e.

Therefore, period of oscillations T = 2π√[(M+m)/(mg/e)] = 2π√[(M+m)e/mg], as given in option (a). (3) The period of vertical oscillations of a mass M suspended using a light spring of spring constant k is T. The same spring is cut into three equal parts and they are used in parallel to suspend the mass M as shown in the adjoining figure. What is the new period of oscillations?

(a) T

(b) 3T

(c) 9T

(d) T/9

(e) T/3

The original period of oscillation T is given by

T = 2π√(M/k)

When the spring is cut into three equal parts, each piece has spring constant 3k.

[Since the length of each piece is reduced by a factor three, the extension for a given applied force will be reduced by a factor three so that the spring constant (which is the ratio of force to extension) will become three times].

Since the three pieces are connected in parallel the effective spring constant of the combination is 3k+3k+3k = 9k. The new period of oscillations T1 is given by

T1 = 2π√(M/9k) = T/3 since T = √(M/k)

You will find some multiple choice questions with solution in this section here.

1. sir I did not understand the last question when the springs are in parallel shouldnt the equivalent spring constant be k/3?

2. Let us consider two springs of spring constants k1 and k2 connected in parallel. When a force F is applied to the combination of springs, the forces F1 and F2 applied on individual springs are given by
F1 = k1x and
F2 = k2x where x is the extension.
The applied force F is the sum of the above forces. Therefore we have
F = keq x = F1 + F2 where keq is the effective spring constant of the combination.
Therefore, keq x = k1x + k2x
This gives keq = k1 + k2
If there are n springs we have
keq = k1 + k2 + k3 +……. kn
In the present case we have 3 identical springs each of spring constant 3k,
Therefore keq = 3k + 3k +3k =9k