A) 1s
B) 2s
C) 3s
D) 4s
Correct Answer: B
Solution :
When 700 g mass is removed, the remaining masses (500 + 400)g oscillate a period of 3 s \[\therefore \] \[3=2\pi \sqrt{\left( \frac{500+400}{k} \right)}\] ?(i) If 500g mass is also removed, the remaining mass is 400 g, its time period \[t=2\pi \sqrt{\frac{400}{k}}\] ?(ii) Dividing Eq. (i) by Eq. (ii) \[\frac{3}{t}=\sqrt{\frac{900}{400}}=\frac{3}{2}\Rightarrow t=2\,s\]You need to login to perform this action.
You will be redirected in
3 sec