# Vector of a particle moving in space help

## Homework Statement

The position vector of a particle moving in space is given by
$$\vec{r}$$=$$(1+cos2wt)$$$$\hat{i}$$ $$+$$ $$3sin^{2}wt$$$$\hat{j}$$$$+$$$$3t$$$$\hat{k}$$
in the ground frame. All the units are in SI. Find the position vector and amplitude of the particle in a frame S moving along the positive z-axis with a velocity 3m/s. Is the line of S.H.M. of the particle parallel to $$4/5$$$$\hat{i}-3/5$$$$\hat{j}$$?

Attempt:

The P.V. of the frame S at any time t is 3t $$\hat{k}$$.
From this we can find out the P.V. of the particle w.r.t. the frame S.

I differentiated it w.r.t. time to find out the velocity vector and found Vmax, which is 5w.
Vmax=Aw
so A=5m

Unfortunately, I got both of them incorrect :/

ehild
Homework Helper

Is it the formula?

$$\vec {r}=(1+\cos(2\omega t)\hat {i }+3\sin^2(\omega t)\hat {j}+3t\hat{k}$$,

(click on it and see the Latex code).

How did you get the result vmax=5w?

The amplitude can be found without the velocity. Find the direction of the SHM and write it along this direction as A(cos2wt)*(unit vector).

ehild

Last edited:

Correction- The component of the P.V. along x-axis is 1+2cos2wt.
First I found out the velocity, then differentiated it to find the condition of Vmax and got wt = pi/4 (one of the solutions)
Subsitituting the value of wt in the equation for velocity, I found out the magnitude of Vmax as 5w.

How to find out the direction of SHM?

ehild
Homework Helper

The velocity is a vector. Only the magnitude of a vector has maximum or minimum.
A simple harmonic motion always happens along a certain direction. On a table, it is horizontal, a body suspended on a spring oscillates vertically, a body on a slope, connected to a spring oscillates along the slope. The amplitude includes information about the direction of oscillation. In general, it is a vector. Think of the polarization of electromagnetic waves.

Collect the cosine terms: you get a cosine multiplied by a vector: it is the vector amplitude.

ehild

The velocity is a vector. Only the magnitude of a vector has maximum or minimum.

Magnitude of the vector has max/min when each of its component has max/min. That happens when wt=pi/4.

Collect the cosine terms: you get a cosine multiplied by a vector: it is the vector amplitude.

ehild
I understand that there is a line of motion of any S.H.M. How do I 'collect' the cosine terms? I don't get your point here.

ehild
Homework Helper

Magnitude of the vector has max/min when each of its component has max/min. That happens when wt=pi/4.
Think it over.
I understand that there is a line of motion of any S.H.M. How do I 'collect' the cosine terms? I don't get your point here.
Edit: As sin2(ωt)=0.5(1-cos(2ωt)) both components of the position vector have a time-dependent cos(2ωt) term. Factoring this out, you have the sum of a constant vector and an oscillating one.

ehild

Last edited:

Think it over.

I was wrong

As sin2(ωt)=1-cos(2ωt) both components of the position vector have a time-dependent cos(2ωt) term. Factoring this out, you have the sum of a constant vector and an oscillating one.

ehild

I got the oscillating term as (i-3j/2)cos2wt
So the amplitude is (13)1/2/2.
But this is not correct.

ehild
Homework Helper

Well, i made a mistake in the previous post: sin2(ωt)=0.5(1-cos(2ωt)).
You have corrected the first formula in your second post : "Correction- The component of the P.V. along x-axis is 1+2cos2wt."

So the position vector is

(i+3/2j)+(2i+3/2 j) cos(2ωt)

at the end, and the magnitude |(2i+3/2 j)|=5/2.

ehild

ok, so the line of SHM of particle is 2i + 3/2j which is not parallel to the vector given in the question.

Thanks!

ehild
Homework Helper

ok, so the line of SHM of particle is 2i + 3/2j which is not parallel to the vector given in the question.

Thanks!

Oh my! one more mistake. I am tired of gardening :) It is

2i -3/2j. It is parallel with the given vector.

ehild