A body executing SHM complete 120 oscillation per minute having amplitude of 1 oscillation find the velocity and acceleration of particle when it is at a distance 5 cm from the mean position
Answers
The particle starts from a distance of 1 cm from the mean position towards the positive extremity. Find the ... Epoch = α = sin-1(xo/a) = sin-1(1/5) = sin-1(0.2) = 11°32' ... If the amplitude of its oscillations is 2 cm, find the velocity. ... We have vmax = ωa .... The periodic time of a body executing S.H.M. is 2 s.
The question is incomplete, below you will find the missing amplitude.
Concept:
we will use the formula for obtaining velocity and acceleration. The number of occurrences of a repeating event per unit of time is frequency
Given:
body complete 120 oscillation per second, amplitude=6 cm, x= 5cm
Find:
- the velocity of the particle
- the acceleration of the particle
Solution:
As given in the question, body complete 120 oscillations in 1 minute
1 min = 120
∴ 1 sec = 120/60 = 2
∴frequency = 2
ω(angular frequency) = 2πf
= 4π
as the formula for velocity we know
v = ω√a²-x²
= 4π√36-25
= 41.65 cm/s
now, the formula for acceleration
A= -ω²x
= -16π²×5
= - 80π²
the negative sign shows the direction
Hence the velocity and acceleration can be find using the formula
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