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प्रश्न
A wave is represented by the equation
\[y = \left( 0 \text{ cdot 001 mm }\right) \sin\left[ \left( 50 s^{- 1} \right)t + \left( 2 \cdot 0 m^{- 1} \right)x \right]\]
(a) The wave velocity = 100 m s−1.
(b) The wavelength = 2⋅0 m.
(c) The frequency = 25/π Hz.
(d) The amplitude = 0⋅001 mm.
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उत्तर
(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm
\[y = \left( 0 \cdot 001 mm \right) \sin\left[ \left( 50 s^{- 1} \right)t + \left( 2 \cdot 0 m^{- 1} \right)x \right]\]
Equating the above equation with the general equation, we get:
\[y = A\sin\left( \omega t - kx \right)\]
\[\omega = \frac{2\pi}{T} = 2\pi v\]
\[k = \frac{2\pi}{\lambda}\]
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
\[A = 0 . 001 mm\]
\[Now, \]
\[50 = 2\pi\nu\]
\[ \Rightarrow \nu = \frac{25}{\pi} Hz\]
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