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.The two points are within one wavelength. Calculate the wavelength, in cm. [Given $S i n^{- 1} (3 / 4) = 0.85 r a d$ ](Round off to nearest integer)

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answer is 250.

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Detailed Solution

Since, $y = A \sin \frac{2 π}{λ} (v t - x)$
$\Rightarrow \frac{y}{A} = \sin 2 π (\frac{t}{T} - \frac{x}{A})$
In the first case, $\frac{y_{1}}{A} = \sin 2 π (\frac{t}{T} - \frac{x_{1}}{λ})$
Where, $y_{1} = + 6, A = 8; x_{1} = 10 c m$
$\Rightarrow \frac{6}{8} = \sin 2 π (\frac{t}{T} - \frac{10}{λ})$
Similarly, in the second case, we get
$\Rightarrow \frac{4}{8} = \sin 2 π (\frac{t}{T} - \frac{25}{λ})$
From equation (1), we get
$2 π (\frac{t}{T} - \frac{10}{λ}) = \sin^{- 1} (\frac{6}{8}) = 0.85 r a d \Rightarrow \frac{t}{T} - \frac{10}{λ} = 0.14$
Similarly, from Equation (2), we get
$2 π (\frac{t}{T} - \frac{25}{λ}) = \sin^{- 1} (\frac{4}{8}) = \frac{π}{6} r a d \Rightarrow \frac{t}{T} - \frac{25}{λ} = 0.08$
Subtracting Equation (4) from equation (3), we get
$\frac{15}{λ} = 0.06 \Rightarrow λ = 250 c m$