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. s'(t) = (b) When is the velocity 0? (Use n as the arbitrary integer.) t =

a) the velocity of particle is -Aωsin (ωt + δ)

b) the velocity is zero when t =  (nπ - δ) / ω, n ∈ Z

Step-by-step explanation:

Given that;

s = A cos(ωt + δ)

(a)

Find the velocity of the particle at time t

the velocity of the particle is given as; v = ds/dt

so differentiate s = A cos(ωt + δ) on both sides with respect to t

v = ds/dt = d/dt [ A cos(ωt + δ) ]

v = A d/dt cos(ωt + δ)

v = -Asin(ωt + δ) d/dt(ωt + δ)

v = -Aωsin (ωt + δ)

therefore, the velocity of particle is -Aωsin (ωt + δ)

b)

When is the velocity 0?

Now since the velocity will be zero; lets set v = 0

-Aωsin (ωt + δ) = 0

sin (ωt + δ) = 0

ωt + δ = nπ

ωt = nπ - δ

t =  (nπ - δ) / ω, n ∈ Z

Therefore the velocity is zero when t =  (nπ - δ) / ω, n ∈ Z

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier