Physics - W.E.P. Question with Solution | TestHub

Using string constraint
$a_1=\frac{a_2+a_3}{2}$
$\Rightarrow 2a_1=a_2+a_3$
$a_1-a_3=a_2-a_1\Rightarrow \left(\mathrm{C}\right)$is correct.
for other options use$m$equivalent

Equivalent mass
Or Reduced mass
$m_{eq}=\frac{2m_1\times m_2}{m_1+m_2}$(for a two-mass system)
$=\frac{2\left(2m\right)\times m}{2m+m}=\frac{4m}{3}$
$\Rightarrow T=\left(\frac{4m}{3}\right).g$
$\therefore 2T=2.\frac{4mg}{3}=\left(\frac{8m}{3}\right)g$
$\Rightarrow m_{eq}=\frac{8m}{3}$

Using mechanical energy conservation
$\Rightarrow \frac{1}{2}k{x}_0^2=\frac{8mg}{3}{x}_0$
$x_0=\frac{16mg}{3k}\Rightarrow \left(\mathrm{A}\right)$is incorrect
Option$\left(\mathrm{B}\right)\Rightarrow V_{\frac{x_0}{2}}=V_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{x_0}{2}\omega =\frac{x_0}{2}\sqrt{\frac{k}{2m+\frac{8m}{3}}}=\frac{x_0}{2}\sqrt{\frac{3k}{14m}}=g\sqrt{\frac{32}{21k}}$
Option$\left(\mathrm{D}\right)\Rightarrow a_{\frac{x_0}{4}}=\frac{x_0}{4}{\omega }^2=\frac{x_0}{4}\frac{3k}{14m}=\frac{3k{x}_0}{42m}=\frac{8g}{21}$
For a simple harmonic motion$x_0\Rightarrow$maximum displacement from extreme to extreme
$\Rightarrow x_0=2A\Rightarrow A=\frac{x_0}{2}$
and$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{3k}{14m}}$
$\therefore v_{\mathrm{m}\mathrm{a}\mathrm{x}}=A.\omega$
and$a={\omega }^2.x$