A) Its amplitude becomes \[\sqrt{2}\] times of its initial value
B) Its periodic time will be doubled
C) Its potential energy will decrease
D) There will be no change in its motion
Correct Answer: A
Solution :
[a] \[\frac{1}{2}K{{A}^{2}}=KE\] at mean position. Now, \[K{{E}_{final}}=2KE\] at mean position \[=\frac{1}{2}m{{(\sqrt{2}{{v}_{0}})}^{2}}\] Hence maximum velocity will be \[\sqrt{2}\] times the previous velocity. Hence, new amplitude will also increase \[\sqrt{2}\] times of its initial value.You need to login to perform this action.
You will be redirected in
3 sec