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Laws of vibrations of stretched strings - class-XI

Description: laws of vibrations of stretched strings
Number of Questions: 50
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Tags: waves physics vibrations of stretched strings
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The frequency of vibration of a sonometer wire is directly proportional to linear density of the wire:

  1. True

  2. False


Correct Option: B
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt{(T/\mu)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

The tension in a piano wire is $10 N$. The tension in a piano wire to produce a node of double frequency is

  1. $20 N$

  2. $40 N$

  3. $10 N$

  4. $120 N$


Correct Option: B
Explanation:

For frequency of oscillation of wire $n \propto \sqrt{T}$. 

Here $T$ is tension in the wire. In order to increase frequency twice, tension needs to be made $4$ times. 
So, new tension must be $4 \times 10 = 40 N$

A knife edge divides a sonometer wire in two parts which differ in length by 2 mm. The whole length of the wire is 1 meter. The two parts of the string when sounded together produce one beat per second. Then the frequency of the smaller and longer pans.in Hz,are

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5


Correct Option: D

A sonometer wire of length $l _1$ vibrates with a frequency 250 Hz. If the length of wire is increased then 2 beats/s are heard. What is ratio of the lengths of the wire?

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57


Correct Option: A
Explanation:

The frequency of sonometer wire is given by
$n=\dfrac{p}{2l}\sqrt{(\dfrac{T}{m})}$
or $n\propto \dfrac{1}{l}$     ...(i)
$\therefore \dfrac{n _1}{n _2}=\dfrac{l _2}{l _1}$
or $\dfrac{250}{250-2}=\dfrac{l _2}{l _1}$
or $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

The tension in the sonometer wire is decreased by 4% by loosening the screws. It fundamental frequency

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary


Correct Option: C
Explanation:

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

The fractional change in frequency with change in tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)

The sonometer wire is vibrating in the second overtone. The length of the wire in terms of wavelength is:

  1. $L=\lambda/2$

  2. $L=5\lambda/2$

  3. $L=3\lambda/2$

  4. $L=3\lambda$


Correct Option: C
Explanation:

second overtone = third harmonic.
Thus, the length of the wire is $L=3\lambda/2$

The correct option is (c)

The frequency of vibration of a sonometer wire is inversely proportional to tension in the wire

  1. True

  2. False


Correct Option: B
Explanation:

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

Thus, frequency of vibration is directly proportional to $\sqrt(T)$

A wire has frequency f. Its length is doubled by stretching. Its frequency now will be:

  1. $1.4\ f$

  2. $0.7\ f$

  3. $2\ f$

  4. $f$


Correct Option: B
Explanation:

$We\quad have\quad f=\frac { 1 }{ 2l } \sqrt { \frac { T }{ m }  } \ \ \therefore f\propto \frac { 1 }{ l } \quad \quad and\quad f\propto \frac { 1 }{ \sqrt { m }  } \ \ And\quad we\quad know\quad that\quad as\quad l\quad becomes\quad double\quad mass/unit\quad length\quad becomes\quad half.\ \ \therefore f\quad becomes\quad \frac { 1 }{ \sqrt { 2 }  } times\quad the\quad orignal.\ \ Hence\quad { f }^{ ' }=0.7f$

Fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubled keeping the tension same, then the new fundamental frequency is :

  1. $\dfrac{n}{2\sqrt{2}}$

  2. $\sqrt{2}n$

  3. $\dfrac{n}{4}$

  4. $\dfrac{2n}{\sqrt{2}}$


Correct Option: C
Explanation:
Frequency of vibration of sonometer wire is given as
$n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}=\dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2d}}\Rightarrow n\propto \dfrac{1}{\sqrt{(d)}}$
If the length and diameter of the wire are doubled. The new frequency will be $\dfrac{n}{4}$.

A wire with linear density of 3 gm/mm is used as a sonometer wire for producing vibrations of frequency 50 Hz. This length of this wire is now halved, while the tension is reduced by 1/4th of the initial tension. What will be the frequency of vibrations produced:

  1. 10 Hz

  2. 30 Hz

  3. 50 Hz

  4. 70 Hz


Correct Option: C
Explanation:

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

since length is halved and tension is made 1/4th the initial tension, their ratio $\sqrt(T)/L$ remains constant. Thus, frequency now depends only on linear mass density

Since the same wire is used, linear mass density dosen't change and hence the frequency also remains same

The correct option is (c)

Two identical sonometer wires have a fundamental frequency of $500$ Hz, when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of $5$ beats/sec, when both wires vibrate together?

  1. 3

  2. 4

  3. 5


Correct Option: A
Explanation:

$n\propto V$
and $V\propto \sqrt{T}$
$\Rightarrow n\propto \sqrt{T}$ ..$(1)$
$5$ beats/sec are obtained when the frequency of one become $505$ Hz i.e. percentage increase in frequency is $1\%$
From $(1)$ Percentage increase in $\eta =1\%$
$\Rightarrow \%$ increases in tension $=2\%$
(Note that method is applicable for small changes only).

When the length of the vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is

  1. 100/101

  2. 99/100

  3. 1

  4. 2


Correct Option: C
Explanation:

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

The fractional change in frequency with change in length is given by $\Delta f/f = -\Delta L/L$.

Thus, the frequency decreases by 1%

The correct option is (c)

A brick is hung from a sonometer  wire. If the brick is immersed in oil, then frequency of the wire will 

  1. increase due to buoyancy

  2. decrease

  3. remains unchanged

  4. increase due to viscosity of oil


Correct Option: B
Explanation:

When immersed in oil , the tension in the string decreases due to force of buoyancy.
$\therefore$ as $f\  \alpha\  \sqrt{T}$
frequency decreases  

The frequency of vibration of a sonometer doubles with doubling the length of the wire.

  1. True

  2. False


Correct Option: B
Explanation:

Frequency of vibration is inversely proportional to length. Hence doubling the length will make the frequency to be halved

The tension in a sonometer wire is found to be 90 N if the distance between the bridges is 30 cm. If the distance is reduced to 10 cm, the tension in the wire will be:

  1. 30 N

  2. 9 N

  3. 90 N

  4. 10 N


Correct Option: C
Explanation:

The tension in the wire does not depend on the distance between the bridges and hence it remains constant

The correct option is (c)

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


Correct Option: D

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


Correct Option: A
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 

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


Correct Option: B
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)

$5\ beats/second$ 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. $251\ Hz$


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 

  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}}}
    $


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 

  1. $200\ Hz$

  2. $150\ Hz$

  3. $600\ Hz$

  4. $450\ Hz$


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 

  1. $T _2 = 2T _1$

  2. $T _2 = 4T _1$

  3. $T _2 = T _1$

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


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 

  1. $1\ kg$

  2. $2\ kg$

  3. $8\ kg$

  4. $16\ kg$


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-

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

  2. $20$ $m / s$

  3. $300$ $m / s$

  4. $400$ $m / s$


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

  1. $0.25\space times$

  2. $0.5\space times$

  3. $4\space times$

  4. $2\space times$


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

  1. $10\mbox{%}$

  2. $19\mbox{%}$

  3. $0.19\mbox{%}$

  4. none of these


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 

  1. $4\space kg$

  2. $2\space kg$

  3. $1\space kg$

  4. $0.5\space kg$


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

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

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

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

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


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?

  1. $4:2:1$

  2. $4:3:1$

  3. $6:3:2$

  4. $3:2:1$


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

  1. $0.08\space m$

  2. $0.02\space m$

  3. $0.13\space m$

  4. $0.16\space m$


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 

  1. 1 : 4 : 9 : 16

  2. 1 : 2 : 3 : 4

  3. 16 : 9 : 4 : 1

  4. 4 : 3 : 2 : 1


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.

  1. 0.3 h

  2. 0.8 h

  3. 0.68 h

  4. 0.5 h


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

  1. 1:2:3:4

  2. 16:9:4:1

  3. 1:4:9:16

  4. 4:3:2:1


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

  1. vibrates with frequency of 416 Hz

  2. vibrates with frequency of 208 Hz

  3. vibrates with frequency of 832 Hz

  4. stop vibrating


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

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


Correct Option: A
Explanation:

The correct answer is B


Given below of the solution query

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

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

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

When tension of a string is increased by 2.5 N, the initial frequency is altered in the ratio of 3:2. The initial tension in the string is 

  1. 6 N

  2. 5 N

  3. 4 N

  4. 2 N


Correct Option: D
Explanation:

Let initial tension = T.
Then given that:
$2\nu = \sqrt{\frac{T}{\mu}}$ $2\nu$ is the frequency
Upon increasing the tension by 2.5N, the frequency becomes $3\nu$
$3\nu = \sqrt{\frac{T+2.5}{\mu}}$
Dividing the two equations.
$\frac{3}{2} =\sqrt{ \frac{T+2.5}{T}}$
$\frac{9}{4} = \frac{T+2.5}{T}$
$9T = 4T + 10$
$5T = 10$
$T = 2N$
Option d is correct.

A transverse wave on a string is given by $\displaystyle y=A\sin \left [ \alpha x+\beta t+\frac{\pi }{6} \right ]$ If $\displaystyle \alpha =0.56/cm,\beta =12/sec,A=7.5cm $ then find the displacement and velocity of oscillation at x = 1 cm and t = 1 s is

  1. $\displaystyle 4.6cm,46.5cm: s^{-1}$

  2. $\displaystyle 3.75cm,77.94cm: s^{-1}$

  3. $\displaystyle 1.76cm,7.5cm: s^{-1}$

  4. $\displaystyle 7.5cm,75cm: s^{-1}$


Correct Option: B

A sonometer wire under a tension of 10 kg weight is in unison with tuning fork of frequency 320 Hz. To make the wire vibrate in unison with a tuning fork of frequency 256 Hz, the tension should be altered by 

  1. 3.6 kg decreased

  2. 3.6 kg increased

  3. 6.4 kg decreased

  4. 6.4 kg increased


Correct Option: A
Explanation:

frequency in a sonometer is given as $f= \dfrac{v}{2l}\sqrt{\dfrac{T}{\mu}}$


$\dfrac{f _1}{f _2} = \sqrt{\dfrac{T _1}{T _2}}$

$\dfrac{320}{256} = \sqrt{\dfrac{10\times g}{T _2}}$

$\dfrac{5}{4} = \sqrt{\dfrac{10\times g}{T _2}}$

$T _2= \dfrac{16}{25} \times 10 g \ N$

$T _2 = 6.4 kg$

Tension to be decreased by 3.6 kg. 

A transverse wave is described by the equation $\displaystyle Y=Y _{0}\sin 2\pi \left ( ft-x/\lambda  \right )$. The maximum particle velocity is equal to four times the wave velocity if

  1. $\displaystyle \lambda =\pi Y _{0}/4 $

  2. $\displaystyle \lambda =\pi Y _{0}/2 $

  3. $\displaystyle \lambda =\pi Y _{0} $

  4. $\displaystyle \lambda =2\pi Y _{0} $


Correct Option: B
Explanation:

$y={ Y } _{ 0 }\sin { 2\pi \left( ft-\frac { x }{ \lambda  }  \right)  } $

Maximum particle velocity
${ V } _{ max }=Aw\ \Rightarrow Aw=4{ V } _{ w }(given)\ Aw=4\left( \frac { w }{ k }  \right) \ A=\frac { 4 }{ k } \ A=\frac { 4 }{ { 2\pi  }/{ \lambda  } } \ \Rightarrow \lambda =\frac { 2\pi A }{ 4 } =\frac { \pi A }{ 2 } \ \left[ \lambda =\frac { \pi { Y } _{ 0 } }{ 2 }  \right] $
Hence option (B) is correct

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates at $120 Hz$. The other end passes over a pulley and supports a $1.50 kg$ mass. The linear mass density of the rope is $0.0550 kg/m$. How wavelength and speed  will change if the mass were increased to $3.00 kg$ ?

  1. both decrease  by $\sqrt{5}$ times.

  2. both increase by $\sqrt{2}$ times.

  3. both increase by $\sqrt{5}$ times.

  4. both decrease  by $\sqrt{2}$ times.


Correct Option: B
Explanation:

Since the mass becomes twice, the tension in the wire becomes twice.

Hence the speed becomes $\sqrt{2}$ times since it varies with tension as $\sqrt{T}$.
$v=\lambda\nu$
Thus the wavelength also becomes $\sqrt{2}$ times the initial value.

The velocity of a transverse wave in a stretched wire is $100ms^{-1}$. If the length of wire is doubled and tension in the string is also doubled, the final velocity of  the transverse wave in the wire is

  1. $100 ms^{-1}$

  2. $141.4 ms^{-1}$

  3. $200 ms^{-1}$

  4. $282.8 ms^{-1}$


Correct Option: B
Explanation:

$v=\sqrt{\dfrac{T}{mass\ per\ unit\ length}}=100\ m/s$
when tension is double $ T$ becomes $2T$ and mass per unit length remains same
so, $v _1 = \sqrt{\dfrac{2T}{mass\ per\ unit\ length}}$
$=\sqrt{2}\times\sqrt{\dfrac T{mass\ per\ unit\ length}}$
$=\sqrt{2}\times 100$
$= 141.4 m/s$

If the vibrations of a string are to be increased by a factor of two, then tension in the string should be made

  1. Twice

  2. Four times

  3. Eight times

  4. Half


Correct Option: B
Explanation:

We know that, $\displaystyle n=\frac {1}{2} \sqrt {\frac {T}{m}} \Rightarrow n \alpha \sqrt {T}$
If tension is increased four times, the frequency will become twice.

A sonometer wire, with a suspended mass of $M=1 kg$, is in resonance with a given tuning fork. The apparatus is taken to the moon where the acceleration due to gravity is $\dfrac 16$ that on earth. To obtain resonance on the moon, the value of $M$ should be

  1. $1 kg$

  2. $\sqrt{6}$ kg

  3. $6 kg$

  4. $36 kg$


Correct Option: C
Explanation:

$f \propto \sqrt{T}=\sqrt{mg}$
$\therefore m _{1}g _{1}=m _{2}g _{2}$
$\Rightarrow (1)g=m\left ( \dfrac{g}{6} \right )$
$\Rightarrow m=6kg$

In an experiment, the string vibrates in $4$ loops when $50 \ gm-wt$ is placed in pan of weight $15 \ gm$. To make the string vibrate in $6$ loops the weight that has to be removed from the pan is approximately :

  1. $72 \ gm$

  2. $36 \ gm$

  3. $21 \ gm$

  4. $29 \ gm$


Correct Option: B
Explanation:

$frequency\propto \dfrac{1}{no\ of\ loops}$
$f _{1}:f _{2}=6:4=3:2$
$f\alpha \sqrt{T}$
$\therefore \dfrac{3}{2}=\sqrt{\dfrac{65}{x}}$
$\dfrac{9}{4}=\dfrac{65}{x}$
$x=\dfrac{260}{9}=29\ gms$
$\therefore$ weight to be removed is $36 gms.$

A uniform wire of length 20 m and weighing 5 kg hangs vertically. If g = 10 m $s^{-2}$, then the speed of transverse waves in the middle of the wire is:

  1. 10 m$s^{-1}$

  2. 10$\sqrt{2}$ m $s^{-1}$

  3. 4 m $s^{-1}$

  4. 2 m $s^{-1}$


Correct Option: A
Explanation:
Here $\mu=\dfrac{5}{20}kg/m=\dfrac{1}{4}kg/m$
Tension in the middle of wire,
$T=$ weight of half the wire $=\dfrac{5}{2}\times g=\dfrac{5}{2}\times 10N=25N$
As,$v=\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow v=\sqrt{\dfrac{25}{1/4}}=10m/s$

If the tension in a sonometer wire is increased by a factor of four. The fundamental frequency of vibration changes by a factor of :

  1. 4

  2. (1/4)

  3. 2

  4. (1/2)


Correct Option: C

The length of a sonometer wire $AB$ is $110 \ cm$. The distance at which two bridges should be placed from $A$ to divide the wire into $3$ segments whose fundamental  frequencies are in the ratio of $1:2:3$ ?

  1. $30 \ cm$

  2. $60 \ cm, 30 \ cm,20 \ cm$

  3. $80\ cm$

  4. $40\ cm, 80\ cm$


Correct Option: B
Explanation:

$f _{1}:f _{2}:f _{3}=1:2:3$
$\Rightarrow \dfrac{1}{l _{1}}:\dfrac{1}{l _{2}}:\dfrac{1}{l _{3}}=1:2:3$
$\Rightarrow l _{1}:l _{2}:l _{3}=6:3:2$
$l _{1}=\left ( \dfrac{6}{2+3+6} \right )\times 110=60\ cm$
$l _{2}=\dfrac{3}{11}\times 110=30\ cm$
$l _{3}=\dfrac{2}{11}\times 110=20\ cm$

If n$ _{1},n _{2},n _{3}$ are the three  fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency '$n$' of the string is given by 

  1. $\sqrt{n}=\sqrt{n _{1}}+\sqrt{n _{2}}+\sqrt{n _{3}}$

  2. $\displaystyle \dfrac{1}{\sqrt{n}}=\dfrac{1}{\sqrt{n _{1}}}+\dfrac{1}{\sqrt{n _{2}}}+\dfrac{1}{\sqrt{n _{3}}}$

  3. $n=n _{1}+n _{2}+n _{3}$

  4. ${\dfrac{1}{n}}=\displaystyle \dfrac{1}{n _{1}}+\dfrac{1}{n _{2}}+\dfrac{1}{n _{3}}$


Correct Option: D
Explanation:

Total length of string is $l=l _{1}+l _{2}+l _{3}$
but $f\propto \dfrac{1}{l}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}+\dfrac{1}{f _{3}}$

To increase the frequency by $20\%$, the tension in the string vibrating on a Sonometer has to be increased by

  1. $44\%$

  2. $33\%$

  3. $22\%$

  4. $11\%$


Correct Option: A
Explanation:

frequency increased by $20\%
$$\Rightarrow f^{'}=\dfrac{6}{5}f$
$\therefore \sqrt{T^{'}}=\dfrac{6}{5}\sqrt{T}$
$\sqrt{T^{'}}=\sqrt{\dfrac{144}{100}}T$
$\therefore$ Tension is to be increased by $44\%$

An iron load of $2 kg$ is suspended in air from the free end of a sonometer wire of length one meter. A tuning fork of frequency $256 Hz$ is in resonance with $1/\sqrt{7}$ times the length of the sonometer wire. If the load is immersed in water, the length of the wire in meter that will be in resonance with the same tuning fork is :


(Specific gravity of iron $= 8$)

  1. $\sqrt{8}$

  2. $\sqrt{6}$

  3. $\displaystyle \dfrac{1}{\sqrt{6}}$

  4. $\displaystyle \dfrac{1}{\sqrt{8}}$


Correct Option: D
Explanation:

$f\propto \dfrac{1}{l}\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow l\propto \sqrt{T}$
$\therefore \dfrac{l _{1}}{l _{2}}=\sqrt{\dfrac{T _{1}}{T _{2}}}=\sqrt{\dfrac{8}{7}}$
$\dfrac{1}{\sqrt{7}l _{2}}=\dfrac{\sqrt{8}}{\sqrt{7}}$
$\Rightarrow l _{2}=\dfrac{1}{\sqrt{8}}$

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