A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is cm/s.
Answer
12
Solution
vmax = A = A 2 T
= 0.06 2 3.143.14 = 0.12 m/s
= 12 cm/s
Answer: 12
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is cm/s.
12
vmax = A = A 2 T
= 0.06 2 3.143.14 = 0.12 m/s
= 12 cm/s
Answer: 12
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