First slide

Simple harmonic motion

Question

A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest it travels a distance $a$ and in the next second it travels a distance $b$ in the same direction. The amplitude of the motion is

Moderate

A

$\frac{2 a^{2}}{3 b - a}$
B

$\frac{3 a^{2}}{3 a - b}$
C

$\frac{2 a^{2}}{3 a - b}$
D

$\frac{3 a^{2}}{3 b - a}$

Solution

Let the acceleration be f; $f = - ω^{2} x$
Therefore, distance of the particle from the centre at any time t is given by
$x = r \cos (ω t)$ where r is the amplitude
$when t = 1 s, x = r - a$
$\begin{array}{l} ∴ (r - a) = r \cos ω \\ \cos ω = \frac{r - a}{r} . . . . . . . . (i) \end{array}$
$When t = 2 s, x = r - a - b,$
$therefore r - a - b = \cos 2 ω$
$∴ r - a - b = r (2 \cos^{2} ω - 1)$ ………..(ii)
Substituting the value of $\cos ω$ from Eq. (i) in Eq. (ii),
$we get r - a - b = r [2 \frac{(r - a)^{2}}{r^{2}} - 1] = \frac{2 (r - a)^{2}}{r} - r$
$∴ r (3 a - b) = 2 a^{2} \Rightarrow r = \frac{2 a^{2}}{3 a - b}$

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