An open pipe of the length $L$ and another open pipe of length $(L + x)$ are sounded together. If the velocity of sound is $V$ and $x << L,$ then the beats frequency produced will be

\frac{V x^{2}}{2 L}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{V L}{2 x^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{V x}{2 L^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{V L^{2}}{2 x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{V x}{2 L^{2}}$
Working formula,
$f_{1} = \frac{V}{2 l}$
$f_{2} = \frac{V}{(2 l + x)}$
Thus, beat frequency,
$f_{1} - f_{2} = \frac{V}{2 l} - \frac{V}{(2 l + x)} = \frac{x V}{2 l (l + x)}$
for very small $x$ ,
$f_{1} - f_{2} = \frac{x V}{2 l^{2}}$

Suggest Corrections

Similar questions

Q. An open pipe of length l is sounded together with another open organ pipe of length

l + x

in their fundamental tones. Speed of sound in air is v. The beat frequency heard will be

(x << l)

Q. An open organ pipe of length

l

is sounded together with another organ pipe of length

l + x

in their fundamental tones

(x << l)

. The beat frequency heard will be

(

speed of sound is

v) :

An open organ pipe of length l is sounded together with another open organ pipe of length l + x in their fundamental tones. Speed of sound in air is v. The beat frequency heard will be (x ≪ l)

Q. In an open end organ pipe of length

l

, if the velocity of sound is

v

then the fundamental frequency will be :

Q. Two open pipes of length

L

are vibrated simultaneously. If length of one of the pipes is reduced by

y

, then the number of beats heard per second will nearly be

(

if the velocity of sound is

v

and

y < L)

Join BYJU'S Learning Program

An open pipe of the length L and another open pipe of length (L+x) are sounded together. If the velocity of sound is V and x<<L, then the beats frequency produced will be

The correct option is B Vx2L2Working formula,f1=V2lf2=V(2l+x)Thus, beat frequency,f1−f2=V2l−V(2l+x)=xV2l(l+x)for very small x,f1−f2=xV2l2

An open pipe of the length $L$ and another open pipe of length $(L + x)$ are sounded together. If the velocity of sound is $V$ and $x << L,$ then the beats frequency produced will be

The correct option is B $\frac{V x}{2 L^{2}}$
Working formula,
$f_{1} = \frac{V}{2 l}$
$f_{2} = \frac{V}{(2 l + x)}$
Thus, beat frequency,
$f_{1} - f_{2} = \frac{V}{2 l} - \frac{V}{(2 l + x)} = \frac{x V}{2 l (l + x)}$
for very small $x$ ,
$f_{1} - f_{2} = \frac{x V}{2 l^{2}}$