Physics - SHM Question with Solution | TestHub

$Y=A\sin \frac{2\pi }{\lambda }\left(vt-x\right)$

$\frac{Y}{A}=\sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$

In first case,$\frac{Y_1}{A}=\sin 2\pi \left(\frac{t}{T}-\frac{x_1}{\lambda }\right)$

Here,$Y_1=+ 6,A=8,x_1=10\mathrm{ }\mathrm{cm}$

$\frac{6}{8}=\sin 2\pi \left(\frac{t}{T}-\frac{10}{\lambda }\right)$.... (i)

Similarly in the second case,

$\frac{4}{8}=\sin 2\pi \left(\frac{t}{T}-\frac{25}{\lambda }\right)$... (ii)

From equation (i),

$\mathrm{2\pi }\left(\frac{t}{T}-\frac{10}{\lambda }\right)={\sin }^{-1}\left(\frac{6}{8}\right)=0.85\mathrm{ }\mathrm{rad}$

$\Rightarrow$$\frac{t}{T}-\frac{10}{\lambda }=0.14$.... (iii)

Similarly from equation (ii),

$2\pi \left(\frac{t}{T}-\frac{25}{\lambda }\right)={\sin }^{-1}\left(\frac{4}{8}\right)=\frac{\pi }{6}\mathrm{ }\mathrm{rad}$

$\Rightarrow \frac{t}{T}=\frac{25}{\lambda }+0.08$.... (iv)

Subtracting equation (iv) from equation (iii), we get

$\frac{15}{\lambda }=0.06$

$\Rightarrow \lambda =250\mathrm{ }\mathrm{cm}$