Tag: sonometer and laws of transverse vibrations

Questions Related to sonometer and laws of transverse vibrations

The tension in a wire is decreased by $19\mbox{%}$. The percentage decrease in frequency will be

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $10\mbox{%}$

  2. $19\mbox{%}$

  3. $0.19\mbox{%}$

  4. none of these

Show answer
Correct Option: A
Explanation:

$\quad f \propto \sqrt T$
$f _{new} = \sqrt{0.81T} = 0.9\sqrt{T}$, that is
 or decrease of $ 10\mbox{%}$

If we add $8\space kg$ load to the hanger of a sonometer. The fundamental frequency becomes three times of its initial value. The initial load in the hanger was about 

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $4\space kg$

  2. $2\space kg$

  3. $1\space kg$

  4. $0.5\space kg$

Show answer
Correct Option: C
Explanation:

$V=\sqrt { \dfrac { TL }{ m }  }$

$T:$tension

$m:$string mass

$L:$string length

$f=\dfrac { V }{ 2L } $          ----- fundamental frequency

suppose initial mass hanging is $M.$

$T=Mg$

${ f } _{ 1 }=\dfrac { V }{ 2L } $

${ f } _{ 1 }=\dfrac { 1 }{ 2L } \sqrt { \dfrac { MgL }{ m }  } $

$final \ \  mass=(M+8)$

${ f } _{ 2 }=\dfrac { 1 }{ 2L } $$\sqrt { \dfrac { (M+8)gL }{ m }  } $

 ${ f } _{ 2 }={ 3f } _{ 1 }$

 

On solving, we get

$M=1$ kg

A uniform rope of length $l$ and mass $M$ hangs vertically from a rigid support. A block of mass $m$ is attached to the free end of the rope. A transverse pulse of wavelength $\lambda$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $\displaystyle \lambda \sqrt{\frac{M - m}{m}}$

  2. $\displaystyle \lambda \frac{M - m}{m}$

  3. $\displaystyle \lambda \sqrt{\frac{m}{M + m}}$

  4. $\displaystyle \lambda \sqrt{\frac{M + m}{m}}$

Show answer
Correct Option: D
Explanation:

the frequency at end and at the begining will remain same.
if one wave is generated at bottom in one sec, only one will reach the top.
hence we need equate frequency at top and bottom
$ \dfrac {{v} _{1}}{{\lambda} _{1}} = \dfrac {{v} _{2}}{{\lambda} _{2}} $
$ {v} _{2}\    \alpha \  \sqrt{tension\   of\   rope } \   \alpha    \sqrt{m + M} $
$ {v} _{1}\    \alpha \  \sqrt{tension\   of \  rope } \   \alpha    \sqrt{m } $

$\implies \dfrac{\lambda _2}{\lambda _1} = \dfrac{v _2}{v _1} = \sqrt {\dfrac{M+m}{m}}$
$\implies \lambda _2= \lambda \sqrt {\dfrac{M+m}{m}}$

A sonometer wire is to be divided in to three segments having fundamental frequencies in the ratio $1:2:3$. What should be the ratio of lengths?

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $4:2:1$

  2. $4:3:1$

  3. $6:3:2$

  4. $3:2:1$

Show answer
Correct Option: C
Explanation:

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
i.e. , for same sonometer, for different segments of bridges length will be the tuning parameter.
 we are given, 
$f _1:f _2:f _3=1:2:3$
while,
$f _1:f _2:f _3=\dfrac{1}{L _1}:\dfrac{1}{L _2}:\dfrac{1}{L _3}$
therefore $L _1:L _2:L _3=\dfrac{1}{f _1}:\dfrac{1}{f _2}:\dfrac{1}{f _3}=6:3:2$

The length of strings of a cello is $0.8\space m$. In order to change the pitch in frequency ratio $5/4$, their length should be decreased by

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $0.08\space m$

  2. $0.02\space m$

  3. $0.13\space m$

  4. $0.16\space m$

Show answer
Correct Option: D
Explanation:

Frequency $\nu \propto \dfrac{1}{l}$. So, $\dfrac{\nu}{\nu'}=\dfrac{l'}{l}$
$\Rightarrow 1-\dfrac{l'}{l}=1-\dfrac{\nu}{\nu'}=1-\dfrac{4}{5}=0.2$
$\Rightarrow \Delta l=0.2\times l =0.16m$ 

Four wires of identical lengths, diameters and materials are stretched on a sonometer wire. The ratio of their tensions is 1 : 4 : 9 : 16. then, the ratio of their fundamental frequencies is 

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. 1 : 4 : 9 : 16

  2. 1 : 2 : 3 : 4

  3. 16 : 9 : 4 : 1

  4. 4 : 3 : 2 : 1

Show answer
Correct Option: B
Explanation:

Fundamental frequency on the wire, $\nu = \dfrac{1}{2  L} \sqrt{\dfrac{T}{\mu}}$   

where, $\mu,   T$ and $L$ are the mass per unit length,  tension and length of the wire respectively.
Now for identical lengths, diameter and materials, $\nu \propto  \sqrt{T}$
Thus, $\nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  \sqrt{T _1}  :  \sqrt{T _2}  :  \sqrt{T _3}   :  \sqrt{T _4}$
  $\nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  \sqrt{1}  :  \sqrt{4}  :  \sqrt{9}   :  \sqrt{16}$
 $\implies      \nu _1  :  \nu _2  :  \nu _3 :  \nu _4   =  1  :  2  :  3   :  4$

A is point on a sonometer wire of uniform area and length L, such that the distances of A from the left end of the wire is $\dfrac { L }{ 18 } $ Find the amplitudes of vibration of the points A if the wire is set vibrating with maximum amplitude h in its ${ 3 }^{ rd }$ harmonic.

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. 0.3 h

  2. 0.8 h

  3. 0.68 h

  4. 0.5 h

Show answer
Correct Option: C

Four wires of identical lengths, diameters and of the same material are stretched on sonometer wire. The ratio of their tensions is 1 : 4 : 9 : 16. The ratio of their fundamental frequencies is

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. 1:2:3:4

  2. 16:9:4:1

  3. 1:4:9:16

  4. 4:3:2:1

Show answer
Correct Option: A

A sonometer wire of length l vibrates in fundamental mode when excited by a tunning fork of frequency 416 Hz. If the length is doubled keeping other things same, the string will

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. vibrates with frequency of 416 Hz

  2. vibrates with frequency of 208 Hz

  3. vibrates with frequency of 832 Hz

  4. stop vibrating

Show answer
Correct Option: A
Explanation:

since the wire is being excited by the tunning fork of frequency $ 416 $ Hz then wire of sonometer will always vibrate at frequency $ 416 $Hz the change in length will only effect its fundamental frequency. 

so the answer is A. 



A transverse wave of amplitude 0.50m, wavelength 1m and frequency 2 Hz is propagating on a string in the negative x direction  The expression form of the wave is

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $\displaystyle y\left ( x,t \right )=0.5\sin \left ( 2\pi x-4\pi t \right )$

  2. $\displaystyle y\left ( x,t \right )=0.5\cos \left ( 2\pi x+4\pi t \right )$

  3. $\displaystyle y\left ( x,t \right )=0.5\sin \left ( \pi x-2\pi t \right )$

  4. $\displaystyle y\left ( x,t \right )=0.5\cos \left ( 2\pi x-2\pi t \right )$

Show answer
Correct Option: A
Explanation:

The correct answer is B


Given below of the solution query

$y(x,t)=Asin(kx+wt)=0.5cos(\dfrac{2\pi}{\lambda }x+2\pi ft)$

$\Rightarrow y(x,t)=0.5cos(\dfrac{2\pi}{1}x+2\pi\times2t)$

$\Rightarrow y(x,t)=0.5cos(2\pi x+4\pi t)$