SHMEasy
Question
An object of mass 0.2 kg executes simple harmonic motion along x-axis with frequency of 25/π Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to :-
Options
A.0.05
B.0.06
C.0.01
D.0.03
Solution
E = 
mω2A2 = 
m(2πf)2A2
∴ A =
Putting E = K + U, we get
A =
= 0.06 m
mω2A2 = m(2πf)2A2∴ A =
Putting E = K + U, we get
A =
= 0.06 mCreate a free account to view solution
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