A free particle is executing simple harmonic motion in a straight line with a time period of 25 seconds; 5 seconds after it has crossed the equilibrium point, the velocity is found to be 0.7m/s. Find the displacement at the end of 10 seconds and also amplitude of oscillations.
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Explanation:
Answer
ω=
T
2π
=
16
2π
=
8
π
rad/s
If at t=0, particle passes through its mean position (x=Asinωt) with maximum speed its v−t equation can be written as
v=ωAcosωt
Substituting the given values, we have
2=(
8
π
)Acos(
8
π
)(2)
∴ A=
π
16
2
m=7.2 m
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