Question

A particle performs SHM 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 $b$ on the same side of the mean position. The amplitude of the SHM is

• A. $a-b$
• B. $\frac{2a-b}{3}$
• C. $\frac{2a^2}{3a-b}$
• D. None of these

Answer 1

Answer: (C)

$\frac{2a^2}{3a-b}$
Particle starts from rest. Hence, we can say that,
$x=A\sin (\omega t+\frac{\pi }{2})=A\cos \left(\omega t\right).......\left(1\right)$
At $t=0,x=A$
Distance travelled by the particle in one second is given by
$a=A-A\cos (\omega \times 1)$
$\Rightarrow \cos \omega =\frac{A-a}{A}=(1-\frac{a}{A})......\left(1\right)$
Distance travelled by the particle in two seconds is given by
$a+b=A-A\cos (\omega \times 2)$
$\Rightarrow a+b=A-A(2{\cos }^2\omega -1)=2A-2A{\cos }^2\omega$
Using $\left(1\right)$ in the above equation, we get
$a+b=2A-2A{(1-\frac{a}{A})}^2$
$\Rightarrow a+b=2A(1-(1-\frac{a}{A}))(1+1-\frac{a}{A})$
$[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$
$\Rightarrow a+b=2A\times \frac{a}{A}(2-\frac{a}{A})$
$\Rightarrow a+b=2a[2-\frac{a}{A}]$
$\Rightarrow \frac{a}{A}=2-\frac{a+b}{2a}$
$\Rightarrow A=\frac{2a^2}{3a-b}$
Thus, option is the correct answer.