Tag: waves

Quizzes (32) Questions (233)

Questions Related to waves

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

A string vibrates in n loops, when the linear mass density is w gm/cm. If the string should vibrate in (n+2) loops, the new wire should have linear mass density:

  1. less than w

  2. more than w

  3. equal to w

  4. equal to w/2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

We also know that frequency is proportional to 1/ number of loops (n)

Thus, $n \alpha \sqrt(\mu)$

Larger n, larger should be the value of $\mu$

The correct option is (b)

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves 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:

  1. $195\ Hz$
  2. $150\ Hz$
  3. $300\ Hz$
  4. $251\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,
$v=195Hz$


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

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 

  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}}}
    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency f is proportional to sqrt(T)/L. In air, f = k*sqrt(Mg)/L1. In water, f = k*sqrt((M-m)g)/L2. Equating the two gives the ratio of weights, which leads to the specific gravity formula D/(D-d) = (L1/L2)^2.

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

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 

  1. $200\ Hz$
  2. $150\ Hz$
  3. $600\ Hz$
  4. $450\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

$f=\dfrac{c}{2l}=\dfrac{300}{1.5}=200Hz$
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

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

  1. $T _2 = 2T _1$
  2. $T _2 = 4T _1$
  3. $T _2 = T _1$
  4. $T _2 = \displaystyle\frac{T _1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 

  1. $1\ kg$
  2. $2\ kg$
  3. $8\ kg$
  4. $16\ kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Fundamental frequency f = (1/2L) * sqrt(T/m). If f is constant and L is doubled, sqrt(T) must double, meaning T must increase by a factor of 4. Since T is proportional to the load, the load must be 4 * 4kg = 16kg.

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

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-

  1. $100$ $\mathrm { m } / \mathrm { s }$
  2. $20$ $m / s$
  3. $300$ $m / s$
  4. $400$ $m / s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The speed of a transverse wave in a stretched wire is given by v = sqrt(T / mu), where T is stress multiplied by cross-sectional area and mu is mass per unit length (density multiplied by area). Thus, v simplifies to sqrt(stress / density). Plugging in 3.0 x 10^8 and 7.5 x 10^5 gives sqrt(400) = 20 m/s.

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

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

  1. $0.25\space times$
  2. $0.5\space times$
  3. $4\space times$
  4. $2\space times$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 

  1. $4\space kg$
  2. $2\space kg$
  3. $1\space kg$
  4. $0.5\space kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
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