Tag: physics

Questions Related to physics

$5\ beats/second$ are heard when a tuning fork is sounded with sonometer wire under tension, when the length of the sonometer wire is either $0.95\ m$ or $1\ m$. The frequency of the fork will be:

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

  1. $195\ Hz$

  2. $150\ Hz$

  3. $300\ Hz$

  4. $251\ Hz$

Show answer
Correct Option: A
Explanation:

When length is 0.95m

$v _1=\frac{v}{2\times 0.95}=\frac{v}{1.9}$
When length is 1m
$v _2=\frac{v}{2\times 1}=\frac{v}{2}$
$ v _1-v=5\quad v-v _2=5$
$ v _1-v _2=10$
$ \frac{v}{1.9}-\frac{v}{2}=10$
$\frac{0.1v}{3.8}=10$
$v=380m/s$
So, $v _1=200Hz , v _2=190Hz$
Then,
$v=195Hz$


A stone is hung in air from a wire, which is stretched over a sonometer. The bridges of the sonometer are 40 cm apart when the wire is in unison with a tuning fork of frequency 256 Hz. When the stone is completely immersed in water, the length between the bridges is 22 cm for re-establishing unison. The specific gravity of material of stone is 

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

  1. $
    \sqrt {\dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}}
    $

  2. $
    \dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}
    $

  3. $
    \dfrac{{40}}
    {{40 - 22}}
    $

  4. $
    \sqrt {\dfrac{{40}}
    {{40 - 22}}}
    $

Show answer
Correct Option: B

The length of a sonometer wire is $0.75\ m$ and density $9\times 10^3  k/m^3$It can bear a stress of $8.1\times 10^8 N/m^2$ with out exceeding the elastic limit The fundamental frequency that can be produced in the wire,is 

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

  1. $200\ Hz$

  2. $150\ Hz$

  3. $600\ Hz$

  4. $450\ Hz$

Show answer
Correct Option: A
Explanation:
Given, $Length=0.75m,density=9\times 10^3k/m^3,Stress=8.1\times10^8N/m^2$

Let the area of the wire be A.

So, $Stress=8.1\times10^8\Rightarrow Density=\dfrac{mass}{volume},mass=Density\times volume$

$=9\times10^3(0.75\times A)=9\times10^3 l\times A$ Where l is the length

$Mass=6.75\times10^3\times A\Rightarrow C=\sqrt{\dfrac{T}{mass/unit}}=\sqrt{\dfrac{8.1 \times 10^8(A)}{\dfrac{6.75\times10^3\times A}{0.75}}}=300m/s$

$f=\dfrac{c}{2l}=\dfrac{300}{1.5}=200Hz$

The fundamental frequency in a stretched string is $100\space Hz$. To double the frequency, the tension in it must be changed to 

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

  1. $T _2 = 2T _1$

  2. $T _2 = 4T _1$

  3. $T _2 = T _1$

  4. $T _2 = \displaystyle\frac{T _1}{4}$

Show answer
Correct Option: B
Explanation:

Fundamental frequency $\nu \propto \sqrt{T}$.
So, $\dfrac{\nu}{\nu'}=\sqrt{\dfrac{T}{T'}}\Rightarrow \dfrac{T}{T'}=\left(\dfrac{\nu}{\nu'}\right)^2=\dfrac{1}{4}$ 
$\Rightarrow T'=4T$. 

A sonometer wire supports a $4\ kg$ load and vibrates in fundamental mode with a tuning fork of frequency $426\ Hz.$ The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to 

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

  1. $1\ kg$

  2. $2\ kg$

  3. $8\ kg$

  4. $16\ kg$

Show answer
Correct Option: D

The density of the material of a wire used in sonometer is $7.5 \times 10 ^ { 5 } \mathrm { kg } / \mathrm { m } ^ { 3 }$  If the stress on the wire is $3.0 \times 10 ^ { 8 } \mathrm { N } / \mathrm { m } ^ { 2 }$ the speed of transverse wave in the wire will be-

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

  1. $100$ $\mathrm { m } / \mathrm { s }$

  2. $20$ $m / s$

  3. $300$ $m / s$

  4. $400$ $m / s$

Show answer
Correct Option: A

The total mass of a sonometer wire remains constant. On increasing the distance between two bridges to four times, its frequency will become

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

  1. $0.25\space times$

  2. $0.5\space times$

  3. $4\space times$

  4. $2\space times$

Show answer
Correct Option: A
Explanation:

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
if $L'=4L$
$f'=\dfrac{1}{8L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4}\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4} f$
Option "A" is correct.

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 } $

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