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Question

The displacement of a particle executing simple harmonic motion is given by equation y=0.3 sin 20π (t+0.05) where time t is in second and displacement y is in metre. calculate the values of amplitude, time period initial phase and initial displacement of the particle .

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Solution

y=0.3sin20π(t+0.05)
y=0.3sin(20πt+π)
Comparing the given equation with the standard equation y=Asin(ωt+ϕ)
By comparison
ω=20π
phi=π
A=0.3
Thus, amplitude and initial phase is known
Amplitude (A)=0.3m
Initial phase (ϕ)=π
Then, for calculating time period
T=2πω=2π20π=110s=0.1s
Time period (T)=0.1s
Also for initial displacement t=0
y=0.3sin(π)=0
Thus, initial displacement =0.

1194741_1358217_ans_afd0dcfb42d745cdb3d3ce99180d1c69.jpg

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