A particle oscillates simple harmonically along a straight line with period 8 s and amplitude 8 2 m. It starts from mean position then the ratio of the distances travelled by it in the 2 n d second

Question

A particle oscillates simple harmonically along a straight line with period 8 s and amplitude 8 2 ​ m. It starts from mean position then the ratio of the distances travelled by it in the 2 n d second and the first second of its motion is ( sin 4 5 ∘ = 2 ​ 1 ​ ) ( sin 9 0 ∘ = 1 )

TestHub · Accepted Answer

The equation of motion for a particle starting from the mean position is given by $x = A sin (ω t)$ .

Given $T = 8$ s, the angular frequency is $ω = \frac{2 π}{T} = \frac{2 π}{8} = \frac{π}{4}$ rad/s.

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

$D_{1} = x (1) - x (0) = A sin (\frac{π}{4} \times 1) - 0 = \frac{A}{2}$

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

$D_{2} = x (2) - x (1) = A sin (\frac{π}{4} \times 2) - \frac{A}{2} = A sin (\frac{π}{2}) - \frac{A}{2} = A - \frac{A}{2}$

The ratio of the distances travelled in the $2^{n d}$ second and the first second is:

$\frac{D _{2}}{D _{1}} = \frac{A - \frac{A}{2}}{\frac{A}{2}} = \frac{\frac{2 - 1}{2}}{\frac{1}{2}} = 2 - 1$

Answer: $(2 - 1)$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion