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 \[n=\frac{1}{2\pi }\sqrt{\frac{k}{2M}}\]You need to login to perform this action.
You will be redirected in
3 sec