Simple Harmonic Motion and time period

[dpsguy]

A particle follows SHM with its eqn.of motion being

s(x,y)=Asinwt i + 3Asin3wt j

What is its time period? Also find the expression for total mechanical energy wrt time.

I tried it in the following way
V=Awcoswt i + 3Awcos3wt j = Aw[(coswt)^2 +9(cos3wt)^2]^(1/2)
KE=0.5m(Aw)^2[(coswt)^2 +9(cos3wt)^2]
a= -[A(w^2){sinwt i + 9sin3wt j }]
F=ma
PE=F.s = -m[(Aw)^2{(sinwt)^2 + 27(sin3wt)^2}]
Total energy=KE + PE
= m(Aw)^2[0.5{(coswt)^2 +9(cos3wt)^2}- {(sinwt)^2 + 27(sin3wt)^2}]
Is this correct? And how to calculate the time period?

 

[HallsofIvy]

"V=Awcoswt i + 3Awcos3wt j = Aw[(coswt)^2 +9(cos3wt)^2]^(1/2)"

This appears to be saying that
cos wt+ 3cos 3wt= (cos²wt+ 9cos²3wt)^1/2 which is NOT true.
$$(a+ b)^2 \ne a^2+ b^2$$
and
$$a+ b \ne \sqrt{a^2+ b^2}$$.

1. What is the period of cos wt?
2. What is the period of cos 3wt?

3. What is the least common multiple of those?

 

[andrevdh]

Why don't you try the basic SHM formulae for the x and y motion separately - because energy is a scalar quantity you can add these up to get the total energy. That is
$$U_x = \frac{1}{2}kx^2$$
for the potential energy and
$$T_x = \frac{1}{2}mv_x^2$$
for the kinetic energy. Just remember that the
$$\omega$$
differs for the two components and use the basic relation
$$\omega = \sqrt{\frac{k}{m}}$$
Great - if you get the same answer ! It just means you have improved the confidence in your result. Anyway, by graphing the motion for a phase from 0 to pi for
$$\omega t$$
the mass moves from 1 to 2 ... p to 5. Thereby the x motion have completed one oscillation, while the y motion have completed one oscillation at point a, another at b and another at point 5. Henceforth the motion repeats itself. The y motion therefore runs 3x faster than the x motion.

 

[dpsguy]

While adding two vectors A and B which are mutually perpendicular
A^2 +B^2=(A+B)^2. Hence I think I was right while finding the velocity. HallsofIvy.
Also,I think i have figured out the answer , with help from andrevdh. The total energy of particle=41m(Aw)^2=constant.The time period comes out to be 3w.
But can't we find the time period of the particle by differentiating the equation for energy twice and using a=-xw^2? Can someone please tell me how can I find the time period here using this method?

 

[andrevdh]

The period is per definition the amount of time needed to complete one oscillation. The attachment displays the motion for
$$\omega t$$
ranging from
$$0\ \rightarrow 2\pi$$
giving one oscillation in the x direction, but three oscillations in the y direction for the corresponding time. After this (1,2, ...,5) the motion will repeat itself. The angular frequency of the x-motion is
$$\omega$$
while it is
$$3\omega$$
for the y-motion. Clearly the period of the combined motion is therefore the period of the x-motion, which can be obtained from
$$\omega=\frac{2\pi}{T}$$
See also 1,2 and 3 of HallsofIvy above.