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, \[v=\frac{\frac{A}{2}}{t}=\frac{\frac{1}{2}}{\frac{0.5}{12}}=12cm/s\]You need to login to perform this action.
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