First slide

Simple harmonic motion

Question

The period of a particle in SHM is 8 sec. At t = 0, it is at the mean position. The ratio of the distances travelled by it in the first one second and in the 2^nd one second is

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{\sqrt{2}}$
C

$\sqrt{2}$
D

$\sqrt{2} + 1$

Solution

At t = 0, let particle move in +x direction. So, equation of SHM is : $x = A \sin ω t$
Now, $ω = \frac{2 π}{T} = \frac{2 π}{8} = \frac{π}{4}$ So, $x = A \sin (\frac{π t}{4})$
$v = \frac{A π}{4} \cos (\frac{π t}{4})$
Distance travelled in t seconds, $d = \int_{0}^{t} |v| d t$
$W h e n [0 < (\frac{π t}{4}) < \frac{π}{2}] o r [0 < t < 2 \sec]$ $v = \frac{A π}{4} \cos (0 < θ < \frac{π}{2}) \Rightarrow v = + v e$
So, velocity is always positive over the time duration 2 seconds studied.
Distance travelled in first 1 second =
$d_{1} = \int_{0}^{1} \frac{A π}{4} \cos (\frac{π t}{4}) d t = {\frac{A π}{4} \frac{4}{π} \sin (\frac{π t}{4})|}_{0}^{1} = A (\frac{1}{\sqrt{2}})$
Distance travelled in second 1 second =
$d_{2} = \int_{1}^{2} \frac{A π}{4} \cos (\frac{π t}{4}) d t = {\frac{A π}{4} \frac{4}{π} \sin (\frac{π t}{4})|}_{1}^{2} = A (1 - \frac{1}{\sqrt{2}})$
$\frac{d_{1}}{d_{2}} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$

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