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) 3*T*

(c) 9*T*

(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 3*k*.

[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 3*k+*3*k+*3*k* = 9*k*. The new period of oscillations *T*_{1}* *is given by

*T*_{1}_{ }*= *2π√(*M/*9*k*) = ** T/3** since

*T =*2π√(

*M/k*)

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

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

ReplyDeleteLet 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

ReplyDeleteF1 = 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