question_answer

A body is in simple harmonic motion with time period half second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it moves form equilibrium position to half of its amplitude.     JEE Main Online Paper (Held On 19 April 2016)

A) 4 cm/s                  

B) 6 cm/s

C) 12 cm/s                               

D) 16 cm/s

Correct Answer: C

Solution :

Given: Time period, T = 0.5 sec Amplitude, A = 1 cm Average velocity in the interval in which body moves from equilibrium to half of its amplitude, v = ? Time taken to a displacement A/2 where A is the amplitude of oscillation from the mean position ?O? is\[\frac{T}{12}\] Therefore, time, \[t=\frac{0.5}{12}\sec \] Displacement, \[s=\frac{A}{2}=\frac{1}{2}cm\] \[\therefore \]Average velocity, \[v=\frac{\frac{A}{2}}{t}=\frac{\frac{1}{2}}{\frac{0.5}{12}}=12cm/s\]