Questions
A particle executes SHM in a line 4 cm long. Its velocity when passing through the centre of line is 12 cm/s. The period will be
detailed solution
Correct option is B
Length of the line = Distance between extreme positions of oscillation = 4 cm So, Amplitude a=2 cm.also vmax=12 cm/s. ∵vmax=ωa=2 πTa ⇒T=2 πavmax =2×3.14×212 =1.047 secTalk to our academic expert!
Similar Questions
A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is . What must be the least period of these oscillations, so that the object is not detached from the platform
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