Question

A particle moves with SHM in a straight line. In first $\tau$ s, after starting from rest it travels a distance a, and in next $\tau$ s it travels 2a in same direction then

• A. Amplitude of motion is 3a
• B. To!e period of oscillations is $8\tau$
• C. Amplitude of motion is 4a
• D. To!e period of oscillations is $6\tau$

Answer 1

Answer: (D)

To!e period of oscillations is $6\tau$

Let $A$ is the amplitude of motion.

$\begin{array}{c}\left(A-a\right)=A\cos wt..........\left(1\right)\\ \&-\left(3a-A\right)=A\cos 2wt\\ \Rightarrow A-3a=A\cos 2wt\\ =A\left[2{\cos }^{2}wt-1\right]\\ =2A{\cos }^{2}wt-A\\ \Rightarrow 2A-2a=2A{\cos }^{2}wt........\left(2\right)\\ \Rightarrow \frac{{\left(A-a\right)}^2}{\left(2A-3a\right)}=\frac{A}{A^2}\\ \Rightarrow \left(A^2+a^2-2Aa\right)=2A^2-3aA\\ \Rightarrow 2A^2+2a^2-4Aa=2A^2-3aA\\ \Rightarrow 2a^2=aA\\ \Rightarrow A=2a\end{array}$

$\therefore$ form $\left(1\right)$

$\begin{array}{c}a=2a\cos wt\\ \Rightarrow \frac{1}{2}=\cos wt\\ \Rightarrow wt=\frac{\pi }{3}\\ \Rightarrow \frac{2\pi }{T}.t=\frac{\pi }{3}\\ \Rightarrow T=6t\end{array}$

$\therefore$ Option $D$ is correct.