First slide

Simple harmonic motion

Question

Time period of a particle executing SHM is 8 s. At t = 0, it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is

Difficult

A

$\frac{1}{\sqrt{2} + 1}$
B

$\sqrt{2}$
C

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

$\sqrt{2} + 1$

Solution

$Here x = A \sin ω t$
$x_{1} = A \sin \frac{2 π}{8} \times 1 = A \sin \frac{π}{4} = \frac{A}{\sqrt{2}}$
$and x_{2} = A \sin \frac{2 π}{8} \times 2 = A$
Therefore, the distance traveled in 2nd second is
$\begin{array}{l} x_{2}^{'} = x_{2} - x_{1} = A - \frac{A}{\sqrt{2}} = \frac{(\sqrt{2} - 1) A}{\sqrt{2}} \\ ∴ Ratio = \frac{x_{1}}{x_{2}^{'}} = \frac{A / \sqrt{2}}{\frac{(\sqrt{2} - 1) A}{\sqrt{2}}} = \frac{1}{(\sqrt{2} - 1)} \\ = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1) (\sqrt{2} + 1)} = \frac{(\sqrt{2} + 1)}{2 - 1} = (\sqrt{2} + 1) \end{array}$

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