Question

two open organ pipes of length 49 cm and 50 cm sounded together beat at 7 per second the speed of sound in air is

a) 336m/s
b) 343m/s
c) 350m/s
d) 357m/s​

Answer 1

Answer:

first answer is wrong

correct answer:-

n1-n2=v/2[1/L1-1/L2]

. ' . 7=v/2[1/49-1/50]×1/10-2

. ' . v= 7×49=343m/s]

Answer 2

The correct option regarding the speed of sound in air is b) 343m/s.

Given:

Two open organ pipes of length 49 cm and 50 cm sounded together beating at 7 per second.

To Find:

The speed of sound in air.

Solution:

To find the speed of sound in the air we will follow the following steps:

Two types of pipers produce sound one is open and the other is a closed pipe. The difference between both pipes is that a closed pipe is closed at both ends while an open pipe is open at one end.

The frequency of closed pipe is given by the formula =

$\frac{v}{4l}$

The frequency of open pipe is given by the formula =

$n = \frac{v}{2l}$

Here, v is the velocity in the air in m/s and l is the length of the organ pipe in meters.

$l2 = 49 \times {10}^{ - 2}$

$l1 = 50 \times {10}^{ - 2}$

(1 meter = 100 cm)

Now,

To get the beats we subtract the frequency of the two organ pipes.

So,

The formula for beats produced in an open organ pipe when two pipes sound together,

$n2 - n1 = \frac{v}{2 \times 49 \times {10}^{ - 2} } - \frac{v}{2 \times 50 \times {10}^{ - 2} }$

n2 - n1 is beats which are 7 per second.

v is the same in both pipes because of the same medium.

$7 = v( \frac{100}{2 \times 49} - \frac{100}{2 \times 50} ) = v (\frac{ 100 }{98} - \frac{100}{100} ) = \frac{100v}{98} - 1v$

$\frac{100v - 98v}{98} = 7$

$2v = 98 \times 7 = 686$

$v = \frac{686}{2} = 343m {s}^{ - 1}$

Henceforth, the correct option regarding the speed of sound in air is b) 343m/s.

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