A particle performing S.H.M., has a period of 12 s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is { cos ( π /6 ) = 2 3

Question

A particle performing S.H.M., has a period of 12 s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is { cos ( π /6 ) = 2 3 ​ ​ }

TestHub · Accepted Answer

The equation of a particle performing S.H.M. starting from the mean position is given by $x = A sin (ω t)$ .

Given the time period $T = 12$ s, the angular frequency is $ω = \frac{2 π}{T} = \frac{2 π}{12} = \frac{π}{6}$ rad/s.

The distance travelled in the first second ( $t = 0$ to $t = 1$ s) is:

$d_{1} = x (1) - x (0) = A sin (\frac{π}{6} \times 1) - 0 = \frac{A}{2}$

The distance travelled in the second second ( $t = 1$ to $t = 2$ s) is:

$d_{2} = x (2) - x (1) = A sin (\frac{π}{6} \times 2) - A sin (\frac{π}{6} \times 1)$

$d_{2} = A sin (\frac{π}{3}) - A sin (\frac{π}{6}) = \frac{A 3}{2} - \frac{A}{2} = \frac{A ( 3 - 1 )}{2}$

The ratio of the distances travelled in the first and second second is:

$\frac{d _{1}}{d _{2}} = \frac{\frac{A}{2}}{\frac{A ( 3 - 1 )}{2}} = \frac{1}{3 - 1}$

Answer: $\frac{1}{3 - 1}$

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