A) \[1\]
B) \[2\]
C) \[0.5\]
D) \[0.25\]
Correct Answer: C
Solution :
In case (i), springs are connected in parallel, so, effective force constant is \[{{k}_{1}}=k+k=2k\] In case (ii), springs are connected in series so, effective force constant \[{{k}_{2}}\] is given by \[\frac{1}{{{k}_{2}}}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}\] \[\Rightarrow \] \[{{k}_{2}}=\frac{k}{2}\] In case (i), time period \[T{{ & }_{1}}=2\pi \sqrt{\frac{M}{{{k}_{1}}}}=2\pi \sqrt{\frac{M}{2k}}\] and in case (ii)/ time period \[T{{ & }_{2}}=2\pi \sqrt{\frac{M}{{{k}_{2}}}}=2\pi \sqrt{\frac{M}{k/2}}\] \[2\pi \sqrt{\frac{2M}{k}}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{1}{4}}\] \[=\frac{1}{2}=0.5\]
You need to login to perform this action.
You will be redirected in
3 sec
