A) 0.137 s
B) 0.328 s
C) 0.628 s
D) 1.00 s
Correct Answer: C
Solution :
The force constant of the spring, \[k=\frac{F}{x}=\frac{0.5\times 10}{0.2}\] = 25 N/m Now, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation, \[T=2\pi \sqrt{\frac{m}{k}}\] \[=2\pi \sqrt{\frac{0.25}{25}}=0.628\,\,s\]
You need to login to perform this action.
You will be redirected in
3 sec
