A particle executing SHM having amplitude 10cm and time period 8 sec, what is displacement of particle after 4 sec;
A. 5cm
B. -20cm
C. 10cm
D. -5cm
E. -10cm
Answers
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Answer:
Given r=10cm
At t=0,x=5cm
T=6sec
So, w=
T
2π
=
6
2π
=
3
π
sec
−1
At, t=0,x=5cm
So, 5=10sin(w×0+ϕ)=10sinϕ [y=rsinwt]
sinϕ=1/2⇒ϕ=
6
π
∴ Equation of displacement x=(10cm)sin(
3
π
)
At t=4second
x=10sin[
3
π
×4+
6
π
]=10sin[
6
8π+π
]
=10sin(
2
3π
)=10sin(π+
2
π
)=−10sin(
2
π
)=−10
Acceleration a=−w
2
x=−(
9
π
2
)×(−10)=10.9≈0.11cm/sec
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