Question

15.A closed organ pipe has length L.The air in it is vibrating in 3rd overtone with a maximum amplitude of A.Find the amplitude at a distance of L/14 from closed end of the pipe ---------

Answer 1

Answer: $a =\dfrac{A}{\sqrt2}$

Explanation:

Given that,

Length = L

Amplitude = A

Distance x = $\dfrac{L}{14}$

We know that,

The equation of standing wave is

$y = A cos\dfrac{2\pi x}{\lambda}\sin\dfrac{2\pi vt}{\lambda}$

The amplitude is a at $x = \dfrac{L}{14}$

The equation of standing  wave

$a = A\cos\dfrac{2\pi}{\lambda}\dfrac{L}{14}$

We know that,

The wave length is

$\lambda = \dfrac{v}{f}$

Now, the equation is

$a = A\cos\dfrac{\pi f}{v}\dfrac{L}{7}$.....(I)

For close organ pipe,

The modes of vibration is

$f = (2n-1)\dfrac{v}{4L}$

For third overtone n = 4

Now, the mode of vibration is

$f = \dfrac{7v}{4L}$

Put the value of f in equation (I)

$a = A\cos\dfrac{\pi 7v}{v4L}\dfrac{L}{7}$

$a = A\cos\dfrac{\pi}{4}$

$a =\dfrac{A}{\sqrt2}$

Hence, the amplitude at a distance of $\dfrac{L}{14}$ from closed end of the pipe is $a =\dfrac{A}{\sqrt2}$.

Answer 2

Answer:

The amplitude at a distance of L/14 from closed end of the pipe is $\dfrac{A}{\sqrt{2}}$.

Explanation:

We need to calculate the amplitude at $x =\dfrac{L}{14}$

From the theory of standing wave

The general equation is

$y = A\cos\dfrac{2\pi x}{\lambda}\sin\dfrac{2\pi vt}{\lambda}$

The amplitude at x is given by

$a =A\cos\dfrac{2\pi l}{\lambda\ 14}$

$a =A\cos\dfrac{\pi l}{\lambda\ 7}$....(I)

We know that,

The wavelength is equal to the velocity of the wave divided by the frequency of wave.

The wavelength is defined as,

$\lambda=\dfrac{v}{f}$

Where, v = velocity

f = frequency

Put the value of wavelength in the equation (I)

$a=A\cos\dfrac{\pi l\ f}{7\ v}$....(II)

For closed organ pipe

The modes of vibration is given by

$f=(2n-1)\dfrac{v}{4l}$...(III)

For third overtone, n = 4

Put the value of n in equation (III)

$f=(2\times4-1)\dfrac{v}{4l}$

$f=\dfrac{7v}{4l}$

Put the value of f in equation (II)

$a=A\cos\dfrac{\pi l\times7v}{4l\times7\ v}$

$a=A\cos\dfrac{\pi}{4}$

$a=\dfrac{A}{\sqrt{2}}$

Hence, The amplitude at a distance of L/14 from closed end of the pipe is $\dfrac{A}{\sqrt{2}}$.