A simple harmonic wave of amplitude 8 units travels along positive X-axis. At any given instant of time, for…
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 the origin, the displacement is + 4 units. Calculate the wavelength.
Solution
Sol. As, $y = A\sin\frac{2\pi}{\lambda}(vt - x)$ or $\frac{y}{A} = \sin 2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)$
In the first case, $\frac{y_1}{A} = \sin 2\pi\left(\frac{t}{T} - \frac{x_1}{\lambda}\right)$
Here, $y_1 = + 6,\; A = 8\;\text{cm},\; x_1 = 10\;\text{cm}$
$\therefore\; \frac{6}{8} = \sin 2\pi\left(\frac{t}{T} - \frac{10}{\lambda}\right) \qquad \dots(\mathrm{i})$
Similarly, in the second case,
$\frac{4}{8} = \sin 2\pi\left(\frac{t}{T} - \frac{25}{\lambda}\right) \qquad \dots(\mathrm{ii})$
From Eq. (i), we get
$2\pi\left(\frac{t}{T} - \frac{10}{\lambda}\right) = \sin^{-1}\left(\frac{6}{8}\right) = 0.85\;\text{rad}$
or
$\frac{t}{T} - \frac{10}{\lambda} = 0.14 \qquad \dots(\mathrm{iii})$
Similarly, from Eq. (ii), we get
$2\pi\left(\frac{t}{T} - \frac{25}{\lambda}\right) = \sin^{-1}\left(\frac{4}{8}\right) = \frac{\pi}{6}\;\text{rad}$
or
$\frac{t}{T} - \frac{25}{\lambda} = 0.08 \qquad \dots(\mathrm{iv})$
Subtracting Eq. (iv) from Eq. (iii), we get
$\frac{15}{\lambda} = 0.06$
$\therefore \; \lambda = 250\;\text{cm}$
Answer: $\lambda = 250$ cm