To determine
To find: The velocity of the particle at time t.
Answer
The velocity of the particle at time t is −Awsin(wt+δ)_.
Explanation
Given:
The equation of motion of a particle is s=Acos(wt+δ).
Derivative rule:
Constant Multiple Rule:
If c is a constant and f(x) is a differentiable function, then
ddx[cf(x)]=cddx[f(x)] (1)
Formula used: Chain Rule
If g is differentiable at x and f is differentiable at g(x), then the composite function F=f∘g defined by F(x)=f(g(x)) is differentiable at x and F′ is given by the product
F′(x)=f′(g(x))⋅g′(x) (2)
Recall:
If x(t) is the displacement of a particle and the time t is in seconds, then the velocity of the particle is v(t)=dxdt.
Calculation:
Obtain the velocity of the particle at time t.
v(t)=ddt(s(t))=ddt(Acos(wt+δ))
Let g(t)=wt+δ and f(u)=Acosu where u=g(t).
Apply the chain rule as shown in equation (2),
v(t)=f′(g(t))⋅g′(t) (3)
The derivative of f′(g(t)) is computed as follows,
f′(g(t))=f′(u)=ddu(f(u))=ddu(Acosu)=Addu(cosu)
Simplify further,
f′(g(t))=A(−sinu)=−Asinu
Substitute u=wt+δ in the above equation,
f′(g(t))=−Asin(wt+δ)
Thus, the derivative f′(g(t))=−Asin(wt+δ).
The derivative of g(t) is computed as follows,
g′(t)=ddt(wt+δ)=ddt(wt)+ddt(δ)=w+0=w
Thus, the derivative g′(t)=w.
Substitute −Asin(wt+δ) for f′(g(t)) and w for g′(t) in equation (3),
v(t)=−Asin(wt+δ)(w)=−Awsin(wt+δ)
Therefore, the velocity of particle at time t is v(t)=−Awsin(wt+δ)_.
To determine
To find: The time such that the velocity is 0.
Answer
The velocity is 0 when t=nπ−δw, n is an integer.
Explanation
Calculation:
From part (a), the velocity of particle is v(t)=−Awsin(wt+δ).
Since the velocity is zero, v(t)=0.
That is, −Awsin(wt+δ)=0.
If A≠0 and w≠0, then −sin(wt+δ)=0.
sin(wt+δ)=0sin−1(sin(wt+δ))=sin−1(0)wt+δ=sin−1(0)
For n∈ℕ, sin−1(0)=nπ.
wt+δ=nπwt=nπ−δt=nπ−δw
Therefore, the velocity is 0 when t=nπ−δw, n is an integer.