Question

A tuning fork with frequency 800 hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. The speed of sound in air is,

Answer 1

Solution :

For the tube open at one the resoN/Asnce frequencies are nv4l when n is a positive odd interger. If the tuning fork has a frequency v and l1,l2,l3 are the successive lengths of the tube in resoN/Ance with it, we have

nv4l1=v

(n+2)v4l2=v

(n+4)v)4l3=v

giv∈gl3−l2=l2−l1=2v4v=v2v

By the question l_3-l_2=(52.75-31.25)cm=21.50cmandl_2-l_1=(31.25-9.75)m=21.50cm.Thus, v/(2v)=21.50cmor v=2vxx21.50cm=2xx800s^-1xx21.50cm=344ms^-1`

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Answer 2

The speed of sound in air is 344 m/s.

Explanation:

Given that,

length $l_{1}=9.75\ cm$

length $l_{2}=31.25\ cm$

length $l_{3}=52.75\ cm$

Frequency = 800 Hz

The tube open at one the resonance frequencies are $\dfrac{nv}{4l}$ when n is a positive odd integer.

If the tuning fork has a frequency v and l₁, l₂, and l₃ are the successive lengths of the tube in resonance with it,

We have,

For l₁,

$f=\dfrac{nv}{4l_{1}}$

For l₂,

$f=\dfrac{(n+2)v}{4l_{2}}$

For l₃,

$f=\dfrac{(n+4)v}{4l_{3}}$

We need to calculate the difference between l₂ -l₁,

$l_{3}-l_{2}=\dfrac{(n+4)v}{4f}-\dfrac{(n+2)v}{4f}$

$l_{3}-l_{2}=\dfrac{2v}{4f}$

We need to calculate the difference between l₃- l₂

$l_{2}-l_{1}=\dfrac{(n+2)v}{4f}-\dfrac{nv}{4f}$

$l_{2}-l_{1}=\dfrac{2v}{4f}$....(I)

Now, given data

The difference of length is

$l_{3}-l_{2}=52.75-31.25$

$l_{3}-l_{2}=21.5\ cm$

Again, The difference of length is

$l_{2}-l_{1}=31.25-9.75$

$l_{2}-l_{1}=21.5\ cm$....(II)

We need to calculate the speed of sound in air

From equation (I) and (II)

$21.5=\dfrac{2v}{4f}$

$v=2\times800\times21.5\times10^{-2}$

$v=344\ m/s$

Hence, The speed of sound in air is 344 m/s.

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Topic : speed of sound

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