
Two closed organ pipes of length 100 cm and 101 cm produces 16 beats in 20 sec fundamental frequency formula in open when each pipe is sounded in its fundamental mode calculate the velocity of sound
A. $303 {m}/{s}$
B. $332 {m}/{s}$
C. $323.2 {m}/{s}$
D. $300 {m}/{s}$
Answer
574.8k+ views
Hint: It is mentioned in the question that both the pipes are sound in their fundamental mode. Use the formula for open ended pipe giving relation between length, velocity and number of beats per second. Number of beats is the difference between two frequencies. Use this basic information and substitute the values in the formula. This will give the velocity of sound.
Formula used:
$f= \dfrac {nv}{4l}$
$\dfrac
{v}{4{l}_{1}}- \dfrac
{v}{4{l}_{2}}= {f}_{1}- {f}_{2}$
Complete answer:
Given: Number of beats $({f}_{1}- {f}_{2})= \dfrac {16}{20}$
${l}_{1}= 100 cm= 1 m$
${l}_{2}= 101 cm= 1.01m$
For an open-ended pipe, formal for frequency is given by,
$f= \dfrac {nv}{4l}$
Where, f is the frequency
n is the ${N}^{th}$ harmonic
v is the velocity of sound
l is the length of the pipe
Number of beats per second is given by,
$\dfrac
{v}{4{l}_{1}}- \dfrac
{v}{4{l}_{2}}= {f}_{1}- {f}_{2}$
Substituting values in above equation we get,
$\dfrac
{v}{4 \times 1}- \dfrac
{v}{4\times 1.01}= \dfrac {16}{20}$
$\Rightarrow \dfrac {v}{4}- \dfrac {v}{4.04}= \dfrac {4}{5}$
$\Rightarrow \dfrac {v}{4}(\dfrac {1}{1}- \dfrac {1}{1.01})= \dfrac {4}{5}$
$\Rightarrow \dfrac {v}{4} \times 9.9 \times {10}^{-3}= \dfrac {4}{5}$
$\Rightarrow v= \dfrac {4}{5} \times \dfrac {4}{9.9 \times {10}^{-3}}$
$\Rightarrow v= 323.23 {m}/{s}$
Hence, the velocity of sound is $323.23 {m}/{s}$.
So, the correct answer is option C i.e. $323.23 {m}/{s}$.
Note:
Students must take care of the unit conversions. They should convert all the quantities to their S.I. units. If they don’t convert the units this may lead them to a wrong output altogether. The longest standing wave in a pipe of length l with two open ends is called a fundamental or first harmonic. The next longest attending wave is known as the second harmonic.
Formula used:
$f= \dfrac {nv}{4l}$
$\dfrac
{v}{4{l}_{1}}- \dfrac
{v}{4{l}_{2}}= {f}_{1}- {f}_{2}$
Complete answer:
Given: Number of beats $({f}_{1}- {f}_{2})= \dfrac {16}{20}$
${l}_{1}= 100 cm= 1 m$
${l}_{2}= 101 cm= 1.01m$
For an open-ended pipe, formal for frequency is given by,
$f= \dfrac {nv}{4l}$
Where, f is the frequency
n is the ${N}^{th}$ harmonic
v is the velocity of sound
l is the length of the pipe
Number of beats per second is given by,
$\dfrac
{v}{4{l}_{1}}- \dfrac
{v}{4{l}_{2}}= {f}_{1}- {f}_{2}$
Substituting values in above equation we get,
$\dfrac
{v}{4 \times 1}- \dfrac
{v}{4\times 1.01}= \dfrac {16}{20}$
$\Rightarrow \dfrac {v}{4}- \dfrac {v}{4.04}= \dfrac {4}{5}$
$\Rightarrow \dfrac {v}{4}(\dfrac {1}{1}- \dfrac {1}{1.01})= \dfrac {4}{5}$
$\Rightarrow \dfrac {v}{4} \times 9.9 \times {10}^{-3}= \dfrac {4}{5}$
$\Rightarrow v= \dfrac {4}{5} \times \dfrac {4}{9.9 \times {10}^{-3}}$
$\Rightarrow v= 323.23 {m}/{s}$
Hence, the velocity of sound is $323.23 {m}/{s}$.
So, the correct answer is option C i.e. $323.23 {m}/{s}$.
Note:
Students must take care of the unit conversions. They should convert all the quantities to their S.I. units. If they don’t convert the units this may lead them to a wrong output altogether. The longest standing wave in a pipe of length l with two open ends is called a fundamental or first harmonic. The next longest attending wave is known as the second harmonic.
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