Tag: sonometer and laws of transverse vibrations

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

A sonometer wire of length 114 cm is fixed at the both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio 1:3:4?

  1. at 36 cm and 84 cm from one end

  2. at 24 cmand 72 cm from one end

  3. at 48 cm and 96 cm from one end

  4. at 72 cm and 96 cm from one end
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequencies are in ratio 1:3:4, so lengths must be in ratio 1/1 : 1/3 : 1/4, which is 12:4:3. Total parts = 19. Lengths are (12/19)*114 = 72cm, (4/19)*114 = 24cm, and (3/19)*114 = 18cm. Placing bridges at 72cm and 96cm (72+24) creates segments of 72, 24, and 18 cm.

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

If the length of the wire of a sonometer is halved the value of resonant frequency will get:

  1. doubled

  2. halved

  3. four times

  4. eight times
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ f=\dfrac{1}{2e}\sqrt{\dfrac{t}{\mu }}$
$f\alpha \dfrac{1}{l}$
$\therefore \dfrac{f _{1}}{f _{2}}=\dfrac{l _{2}}{l _{2}}=\dfrac{1}{2}$
$\Rightarrow f _{2} = 2f _{1}$
$\therefore $ frequency is doubled 

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
A Correct answer
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.