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Question: If equation of sound wave is \(y = 0.0015\sin (62.4x + 316t)\) , then its wavelength will be (A...

If equation of sound wave is y=0.0015sin(62.4x+316t)y = 0.0015\sin (62.4x + 316t) , then its wavelength will be

(A) 2 unit

(B) 0.3unit

(C) 0.1unit

(D) 0.2 unit

Explanation

Solution

The given equation takes the exact form of the progressive wave equationy=Asin(ωtkx)y = A\sin (\omega t - kx). So, on comparing both the wave functions, find out the value of the propagation constant k. k is related to the wavelength by k=2πλk = \dfrac{{2\pi }}{\lambda } . Now on substituting the values calculate the wavelength of the wave.

Complete step-by-step answer
A mathematical description of the disturbance created by a wave is called a wave function.

We know that, a progressive wave function is given by,
y=Asin(ωtkx)y = A\sin (\omega t - kx)

Where,
Y is the displacement; A is the amplitude; ω\omega is the angular frequency; x is the position; k is the propagation constant and t is the time.
y=0.0015sin(62.4x+316t)y = 0.0015\sin (62.4x + 316t)

The given equation is similar to this wave function and therefore has corresponding values. On comparing with the progressive wave function we find out the value of the propagation constant also known as angular wave number.
k=62.8k = 62.8

The number of wavelengths in the distance2π2\pi is called the wave number which is given by,
k=2πλk = \dfrac{{2\pi }}{\lambda }

Substitute the value of the angular wave number.

λ=2×3.1462.8\lambda = \dfrac{{2 \times 3.14}}{{62.8}}

λ=0.1unit\lambda = 0.1unit

Hence, the wavelength of the wave is 0.1 units and the correct option is C.

Note: If during the propagation of a progressive wave, the particles of the medium perform SHM about their mean position then the wave is known as a harmonic progressive wave. Various forms of progressive wave function.

y=Asin2π(tTxλ)y = A\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right)

y=Asin2πT(txTλ)y = A\sin \dfrac{{2\pi }}{T}\left( {t - \dfrac{{xT}}{\lambda }} \right)

y=Asin2πλ(vtx)y = A\sin \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right)

y=Asinω(txv)y = A\sin \omega \left( {t - \dfrac{x}{v}} \right)

We can also write it in exponential form

y=Aeiω(txv)eiω(txv)2iy = A\dfrac{{{e^{i\omega \left( {t - \dfrac{x}{v}} \right)}} - {e^{ - i\omega \left( {t - \dfrac{x}{v}} \right)}}}}{{2i}}