A) \[\frac{1}{2\pi }\sqrt{\frac{k}{M}}\]
B) \[\frac{1}{2\pi }\sqrt{\frac{2k}{5M}}\]
C) \[\frac{1}{2\pi }\sqrt{\frac{k}{5M}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{2k}{M}}\]
Correct Answer: B
Solution :
Upper two springs are in series. Therefore, their resultant spring constant \[\frac{1}{k},\]\[=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}\] \[\therefore \] \[k'=\frac{k}{2}\] Lower two springs are in parallel. Therefore their resultant spring constant, \[k''=k+k=2k\] Now, springs of constant k' and k'' are in series. Therefore, their resultant spring constant \[\frac{1}{{{k}_{1}}}=\frac{2}{k}+\frac{1}{2k}=\frac{4+1}{2k}\] \[{{k}_{1}}=\frac{2k}{5}\] \[\therefore \] Frequency of oscillation \[f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}}{M}}\] \[=\frac{1}{2\pi }\sqrt{\frac{2k}{5M}}\]You need to login to perform this action.
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