   Chapter 3.4, Problem 80E

Chapter
Section
Textbook Problem

If the equation of motion of a particle is given by s = A cos(ωt + δ), the particle is said to undergo simple harmonic motion.(a) Find the velocity of the particle at time t.(b) When is the velocity 0?

(a)

To determine

To find: The velocity of the particle at time t.

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=fg 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)

(b)

To determine

To find: The time such that the velocity is 0.

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