A) \[2\pi \sqrt{\frac{M}{k}}\]
B) \[2\pi \sqrt{\frac{M}{2k}}\]
C) \[2\pi \sqrt{\frac{M}{4k}}\]
D) \[2\pi \sqrt{\frac{2M}{k}}\]
Correct Answer: C
Solution :
| When mass M is |
 |
| suspended from the given system as shown in figure, let \[l\] be the length through which mass M moves down before it comes to rest. |
| In this situation, both the spring and string will be stretched by length l. Since string is inextensible, so spring is stretched by length\[2l\]. The tension along the string and spring is the same. |
| In equilibrium, \[Mg=2\,(k2l)\] |
| If mass M is pulled down through small distance x, then |
| \[F=Mg-2k(2l+2x)=-4kx\] ...(i) |
| \[F\propto x\] and \[-ve\] sign shows that it is directed towards mean position. Hence, the mass executes simple harmonic motion. |
| For SHM, \[F=-k'x\] b ...(ii) |
| Comparing (i) and (ii), we get \[k'=4k\] |
| \[\therefore \] Time period, \[T=2\pi \sqrt{\frac{M}{4k}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
