When does a Particle Come to Rest and Reach Maximum Velocity During Oscillation?

[steve2510]

Homework Statement

A particle oscillates about a fixed point. Its distance, x(m) from the origin is given by the equation x=3sin(2t) + 2cos(2t) -2.

Find

i) its velocity,
ii) where it first comes to rest,
iii) its maximum velocity.

Homework Equations

The Attempt at a Solution

Well firstly dx/dt = v
∴ v = 6cos(2t) - 4sin(2t)

For i) The velocity is just found out by placing a value of T into the above equation. I'm confused as to why no value of t has been stated as the velocity is changing depending on what value of t is used.

For ii) It comes to rest when v=0
So 6cos(2t) - 4cos(2t) = 0 A=2T

Rcos(A-B) = RcosAcosB - RsinAsinB

Comparing coefficents of cosA and SinA,

RcosB= 6 RsinB= 4

RsinB/RcosB = TanB
B = tan^-1 4/6 = 0.588 rads = B

If RsinB = 4 ...R(sin 0.588) = 4
R= 7.21

7.21(cos(A-0.588) = 6cos(2t) - 4sin(2t) = 0
7.21(cos(A-0.588)) = 0 A=2t
Cos = 0 at pi/2 and 3pi/2 so Cos(2t-0.588) = 0
2t-0.588 = pi/2
2t= 2.159
t= 1.080

If I put that back into the equation, it doesn't equal 0 which means I must have gone wrong somewhere.

iii) max velocity, I assume I need to take the derivative again and plug in values of t to see which produces a maximum i.e <0

 

[ehild]

Homework Statement

A particle ossilattes about a fixed point. Its distance, x(m) from the origin is given by the equation
x=3sin(2t) + 2cos(2t) -2
Find i) its velocity
ii) where it first comes to rest
iii) its maximum velocity

Homework Equations

The Attempt at a Solution

Well firstly dx/dt = v
∴ v = 6cos(2t) - 4sin(2t)
For i) The velocity is just found out by placing a value of T into the above equation, I am confused as to why no value of t has been stated as the velocity is changing depending on what value of t is used.
For ii) It comes to rest when v=0
So 6cos(2t) - 4cos(2t) = 0

It is 6cos(2t) - 4sin(2t)=0 instead. Which means 4sin(2t)=6cos(2t), that is tan(2t)=1.5


ehild

 

[steve2510]

Okay, I can't believe I didn't see that, so at that time, it comes to rest. To find where it comes to rest, I guess I plug that t back into the orignal equation.

Then to find the maximum value of velocity hmm? I'm reading maximum, so I'm thinking second derivative, but I'm not sure how that helps.

 

[milesyoung]

Okay can't believe i didnt see that, so at that time it comes to rest to find where it comes to rest i guess i plug that t back into the orignal equation

Yeah, use your value of t to find x(t), the position of the particle at time t.

Then to find the maximum value of velocity hmm? I am reading maximum so I am thinking second deriavitve but I am not sure how that helps

You've probably done it a lot of times before. Find the roots of dv/dt (solutions to dv/dt = 0) to find values of t that gives maxima/minima of v.