Question

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

• A. 150 cm
• B. 250 cm
• C. 350 cm
• D. 550 cm

Answer 1

Answer: (B)

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