You need to solve
[tex]-a\omega\sin(\omega t+\delta)=0[/tex]
for [tex]t[/tex]. If [tex]a\omega\neq0[/tex], then we must have
[tex]\sin(\omega t+\delta)=0[/tex]
We have [tex]\sin x=0[/tex] for [tex]x=n\pi[/tex], where [tex]n[/tex] is any integer, so
[tex]\omega t+\delta=n\pi\implies t=\dfrac{n\pi-\delta}\omega[/tex]
The periodic motion of the particle consists of velocities that are within a
given range and a domain that extends to infinity.
Reasons:
The given parameters are;
The function for the motion of the particle is; s = a·cos(ω·t + δ)
The velocity of the particle is found as s'(t) = -a·ω·sin(ω·t + δ)
Required:
The time at which the velocity is zero
Solution:
At zero velocity, we have;
The input variable is the time, t, therefore;
sin(ω·t + δ) = 0
From the attached graph, the value of sin(θ) is zero for multiples of π,
therefore, we have;
sin(ω·t + δ) = sin(n·π), where; n is an integer
Which gives;
Making t subject of the above formula gives;
Therefore;
Learn more here:
Simple harmonic motion
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