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Time when they meet again

1 vote

I found the time period in both as $\pi /10$ and $3\pi /10$
If I assume one block at rest then the relative time period would be $\pi /5 $
After that I got confused .

Answer s given as $3\pi /10$

asked Mar 28, 2017 in Physics Problems by koolman (4,286 points)

1 Answer

2 votes
Best answer

The only physics here is finding the period of each oscillator, using $\omega=\sqrt{\frac{k}{m}}$. These are $\omega_1=20 rad/s$ and $\omega_2=\frac{20}{3}rad/s$. The rest is mathematics.

When in difficulty with calculations, you can always sketch or plot a graph (below). The displacements are $x_1=10\cos(20t)$ and $x_2=10\cos(\frac{20}{3}t)$.

My own initial confusion is understanding what is meant by being again side by side. It could mean that the blocks are :
1. passing and moving in the opposite direction; or
2. moving in the same direction , but not necessarily back in the start position; or
3. back at the start position.
These 3 options are illustrated in the graph above.

There is no way of knowing which time is intended. The times for each case are 1. $\frac34 \frac{\pi}{10}$ seconds, 2. $\frac32 \frac{\pi}{10}$ seconds, and 3. $3\frac{\pi}{10}$ seconds.

To get the same results by calculation we set $x_1=x_2$ :
$10\cos(\omega_1 t)=10\cos(\omega_2 t)$
$\omega_1 t = 2\pi n \pm \omega_2 t$
$(\omega_1 \pm \omega_2)t=2\pi n$
where $n$ is an integer, and I have used the fact that the cosine has reflection symmetry $(\cos(-x)=\cos(x))$ as well as periodic translation symmetry $(\cos(x)=\cos(x+2\pi n))$.

Taking the + sign we get for $n=1$ that
Taking the - sign we get
$\frac{40}{3} t=2\pi$
$t=\frac{3 \pi}{20}$.
Taking the - sign with $n=2$ we get
$t=\frac{3 \pi }{10}$.

Alternatively, starting from your calculation of the periods $T_1=\frac{\pi}{10}$ and $T_2=\frac{3\pi}{10}$ :

The blocks will return to the starting position at the same time when each block has made a whole number of cycles. The time after which the blocks reach this position is
The smallest possible integer values are $m=3, n=1$. Then $t=nT_2=\frac{3\pi}{10}$.

answered Mar 28, 2017 by sammy gerbil (28,806 points)
selected Mar 29, 2017 by koolman