question_answer

Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, the frequency of oscillation of block is:

A) \[\frac{1}{2\pi }\sqrt{\frac{k}{M}}\]        

B)       \[\frac{1}{2\pi }\sqrt{\frac{k}{2M}}\]     

C)       \[\frac{1}{2\pi }\sqrt{\frac{2k}{M}}\]     

D)       \[\frac{1}{2\pi }\sqrt{\frac{M}{k}}\]

Correct Answer: B

Solution :

                Let when the oscillating mass is at a distance x towards right from its equilibrium position, the instantaneous extensions in the springs of force constants k, is \[x={{x}_{1}}+{{x}_{2}}\] Since, the springs are in series the restoring force exerted by each spring on mass m is same. Then \[F=-k{{x}_{1}}=-k{{x}_{2}}\] \[\therefore \]   \[{{x}_{1}}=-\frac{F}{k},{{x}_{2}}=-\frac{F}{k}\] and   \[x={{x}_{1}}+{{x}_{2}}=-F\left( \frac{1}{k}+\frac{1}{k} \right)=\frac{-2F}{k}\] \[\therefore \]   \[F=-\frac{k}{2}x\] \[\Rightarrow \]  Effective force constant is \[\frac{k}{2}\]. Hence, frequency of oscillation is \[n=\frac{1}{2\pi }\sqrt{\frac{k}{2M}}\]