Question

A simple harmonic wave of amplitude $8$ unit travels along positive $x$-axis. At any given instant of time, for a particle at a distance of $10cm$ from the origin, the displacement is $+6$ unit and for a particle at a distance of $25cm$ from the origin, the displacement is $+4$ unit. Calculate the wavelength.

• A. $200cm$
• B. $230cm$
• C. $210cm$
• D. $250cm$

Answer 1

The correct option is D $250cm$

$Y=A\sin \frac{2\pi }{\lambda }(vt-x)$
$\frac{Y}{A}=\sin 2\pi (\frac{t}{T}-\frac{x}{\lambda })$
In first case, $\frac{Y_1}{A}=\sin 2\pi (\frac{t}{T}-\frac{x_1}{\lambda })$
Here, $Y_1=+6$, $A=8$, $x_1=10cm$
$\frac{6}{8}=\sin 2\pi (\frac{t}{T}-\frac{10}{\lambda })$ .....(i)
Similarly in the second case,
$\frac{4}{8}=\sin 2\pi (\frac{t}{T}-\frac{25}{\lambda })$ .....(ii)
From equation (i),
$2\lambda (\frac{t}{T}-\frac{10}{\lambda })={\sin }^{-1}\left(\frac{6}{8}\right)=0.85rad$
$\Rightarrow \frac{t}{T}-\frac{10}{\lambda }=0.14$ ....(iii)
Similarly from equation (ii),
$2\pi (\frac{t}{T}-\frac{25}{\lambda })={\sin }^{-1}\left(\frac{4}{8}\right)=\frac{\pi }{6}rad$
$\Rightarrow \frac{t}{T}-\frac{25}{\lambda }=0.08$ ....(iv)
Subtracting equation (iv) from equation (iii), we get
$\frac{15}{\lambda }=0.06$
$\Rightarrow \lambda =250cm$