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 $10cm$ from the origin, the displacement is $ + 6$ units, and for a particle at a distance of $25cm$ from the origin, the displacement is$ + 4$ units. Calculate the wavelength.${u_1} = 30cm$.
$\left( A \right) 200cm$
$\left( B \right) 230cm$
$\left( C \right) 210cm$
$\left( D \right) 250cm$

Answer 1

Hint: Simple harmonic motion is the repetitive back and fro movement through a central position or equilibrium. The time interval of each complete oscillation remains the same. Now write the sine wave equation to both the cases. Subtract the two-equations to obtain the wavelength.

Formula used:
$\dfrac{Y}{A} = \sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$
Where $A$ is the amplitude, $\lambda $ is the wavelength, $Y$ is the displacement.

Complete step by step solution:
Simple harmonic motion is the repetitive back and forth movement through a central position or equilibrium. The time interval of each complete oscillation remains the same.The force responsible for the back and fro movement is directly proportional to the distance between them.$F = - kx$ this relation is the hooke's law.
In simple harmonic motion wavelength is directly proportional to the speed and of sound and inversely proportional to the frequency of a simple harmonic motion. The particles in the medium in most of the periodic waves exhibit simple harmonic motion.
The sine wave equation
$\Rightarrow$ $Y = A\sin \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right)$
$\Rightarrow$ $\dfrac{Y}{A} = \sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)$
Here in the first case$Y = + 6$,$A = 8$,$x = 10cm$
$\Rightarrow$ $\dfrac{6}{8} = \sin 2\pi \left( {\dfrac{t}{T} - \dfrac{{10}}{\lambda }} \right) - - - - - - - \left( 1 \right)$
In the second case,
$\Rightarrow$ $\dfrac{4}{8} = \sin 2\pi \left( {\dfrac{t}{T} - \dfrac{{25}}{\lambda }} \right) - - - - - - \left( 2 \right)$
From equation \[\left( 1 \right)\]
$\Rightarrow$ $\left( {\dfrac{t}{T} - \dfrac{{10}}{\lambda }} \right) = 0.14$
From equation \[\left( 2 \right)\]
$\Rightarrow$ $\left( {\dfrac{t}{T} - \dfrac{{10}}{\lambda }} \right) = 0.08$
Subtract the two equations to get
$ \Rightarrow \dfrac{{15}}{\lambda } = 0.06$
$\therefore$ $\lambda = 250cm$
Hence option $\left( D \right)$ is the right option.

Note: The particle will oscillate along the direction of the wave. The force responsible for the back and fro movement is directly proportional to the distance between them.$F = - kx$ this relation is the hooke's law. The particles in the medium in most of the periodic waves exhibit simple harmonic motion.