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tino4ka555 [31]
3 years ago
13

Two closed organ pipe gives 4 beats when sounded together at 5°C. Calculate the number of beats in 35°C​

Physics
1 answer:
Gre4nikov [31]3 years ago
7 0

Answer:

The number of beats is 10.58 in 35°C.

Explanation:

The beat frequency is given by : f₁-f₂

At 5°C, f₁-f₂ = 4

We need to find the number of beats in 35°C​.

The frequency in a standing wave is proportional to \sqrt T.

So,

\dfrac{f_1-f_2}{f_1'-f_2'}=\dfrac{\sqrt {T}}{\sqrt{T'}}\\\\f_1'-f_2'=(f_1-f_2)\times \dfrac{\sqrt{T'}}{\sqrt{T}}\\\\f_1'-f_2'=4\times \dfrac{\sqrt{35}}{\sqrt{5}}\\\\f_1'-f_2'=10.58

So, the number of beats is 10.58 in 35°C.

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              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

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              t is the time = 0.16 s

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                   v(t) = \frac{dy}{dt}

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The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})


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The period of the wave is

             T = \frac{2 \pi}{\omega}

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                    v = \frac{\lambda}{T}

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therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

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