# 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?

Homework Helper
"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?

Homework Helper
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.

#### Attachments

• 2D SHM.bmp
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Last edited:
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?

Homework Helper
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
$$\omega=\frac{2\pi}{T}$$