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)

$Y=Asin\frac{2\pi }{\lambda }\left(vt-x\right)$
$\frac{Y}{A}=sin2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$
In first case, $\frac{Y_1}{A}=sin2\pi \left(\frac{t}{T}-\frac{x_1}{\lambda }\right)$
Here, $Y_1=+6,A=8,x_1=10cm$
$\frac{6}{8}=sin2\pi \left(\frac{t}{T}-\frac{10}{\lambda }\right)$ .... (i)
Similarly in the second case,
$\frac{4}{8}=sin2\pi \left(\frac{t}{T}-\frac{25}{\lambda }\right)$ ... (ii)
From equation (i),
$2\pi \left(\frac{t}{T}-\frac{10}{\lambda }\right)={\left(sin\right)}^{-1}\left(\frac{6}{8}\right)=0.85rad$
$\Rightarrow$ $\frac{t}{T}-\frac{10}{\lambda }=0.14$ .... (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
$\frac{15}{\lambda }=0.06$
$\Rightarrow \lambda =250cm$