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Standing waves in strings - class-XI

Description: standing waves in strings
Number of Questions: 34
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Tags: oscillation and waves physics oscillations and waves waves
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A string is under tension so that its length is increased by $1/n$ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be

  1. $1:n$

  2. ${n}^{2}:1$

  3. $\sqrt{n}:1$

  4. $n:n+1$


Correct Option: D

Motion that moves to and fro in regular time intervals is called _________________ motion.

  1. Vibratory

  2. Translatory

  3. Rotatory

  4. Accelerating


Correct Option: A
Explanation:

Sounds are made by vibrations. Some vibrations are easy to see. The vibrations that create sound must travel through a medium, such as air or water, or anything made of molecules. With each forward motion, air molecules pulse outward, pushing other air molecules and crowding them together. With each backward motion, the molecules get less crowded. The forward and backward vibration of the glass creates a chain reaction of crowded and not-so-crowded molecules that ripples through the air. This traveling vibration is called a sound wave. 
Motion that moves to and fro in regular time intervals is called vibratory or oscillatory motion.

When we hear a sound, we can identify its source from : 

  1. Amplitude of sound

  2. Intensity of sound

  3. Wavelength of sound

  4. Overtones present in the sound


Correct Option: D
Explanation:

Answer is D.

When we hear a sound, we can identify its source from overtones present in the sound.
The fundamental is the frequency at which the entire wave vibrates. Overtones are other sinusoidal components present at frequencies above the fundamental. All of the frequency components that make up the total waveform, including the fundamental and the overtones, are called partials. Together they form the harmonic series.
Overtones which are perfect integer multiples of the fundamental are called harmonics. When an overtone is near to being harmonic, but not exact, it is sometimes called a harmonic partial, although they are often referred to simply as harmonics. Sometimes overtones are created that are not anywhere near a harmonic, and are just called partials or inharmonic overtones.

The vibrations produced by the body after it is into vibration is called ....................

  1. Force Vibrations

  2. Free or Natural Vibrations

  3. Damped Vibrations

  4. None of these


Correct Option: B

The length of a stretched string is $2 m$. The tension in it and its mass are $10 N$ and $0.80 kg$ respectively. Arrange the following steps in a sequence to find the third harmonic of transverse wave that can be created in the string.
(a) Find the linear mass density ($m$) using the formula, $m$ $\displaystyle = \dfrac{mass (M)  of \ the \  string}{length (l)  of \ the \  string}$
(b) Collect the data from the problem and find the length($l$) tenstion ($T$) and mass ($M$) of the stretched string.
(c) The fundamental frequency of a stretched vibrating string is given by $n$ $=\displaystyle \dfrac{1}{2l} \sqrt{\dfrac{T}{m}}$
(d) The frequency of $2^{nd}$ overtone or $3^{rd}$ harmonic is given by $n _2\displaystyle = \dfrac{3}{2l}\sqrt{\dfrac{T}{m}}=3n$.

  1. a b c d

  2. d b c a

  3. b a c d

  4. b d c a


Correct Option: C
Explanation:

Collect the data from the problem and find the length ($l$), tension ($T$) and mass ($M$) of the stretched string (b). 

Find the linear mass density (m) using the formula, m $= m/l$  (a).
The fundamental frequency of a stretched vibrating string is given by, n $=\displaystyle \dfrac{1}{2l} \sqrt{\dfrac{T}{m}}$ (c).
The frequency of $2^{nd}$ overtone or $3^{rd}$ harmonic is given by $n _2=\displaystyle \dfrac{3}{2l} \sqrt{\dfrac{T}{m}}=3n$ (d)

All overtones are stationary wave.

  1. True

  2. False


Correct Option: B
Explanation:

All overtones are not stationary waves. Only those overtones which match the frequencies of the harmonics act as stationary waves.

All harmonics in a stringed instrument are

  1. Standing waves

  2. Progressive wave

  3. Electromagnetic waves

  4. Transverse waves


Correct Option: A
Explanation:

All harmonics in a stringed instrument are stationary or standing waves

The correct option is (a)

nth overtone and (n/2) harmonic are always equal

  1. True

  2. False


Correct Option: B
Explanation:

$n^{th}$ overtone is always equal to (n+1) harmonic.

For example :
First overtone = second overtone 
Second overtone = third overtone 

All harmonics are possible in a string fixed at one end

  1. True

  2. False


Correct Option: B
Explanation:

In a string fixed at one end, only odd harmonics are allowed

A set of 3 standing waves 5, 10 and 15 Hz are to be setup on a string fixed at one end. One of these frequencies are suppressed, while passing through it. Identify them:

  1. 5 Hz

  2. 10 Hz

  3. 15 Hz

  4. All the frequencies will pass through them


Correct Option: B
Explanation:

In a string fixed at one end, only odd harmonics are allowed and even harmonics are suppressed.

Thus the 10Hz standing wave is suppressed,
The correct option is (b)

The 3rd overtone for a string fixed at one end is 

  1. 3rd harmonic

  2. 5th harmonic

  3. 1st harmonic

  4. fundamental note


Correct Option: B
Explanation:

The  third overtone is the 5th harmonic, since odd harmonics are only allowed in  a string fixed at one end

The correct option is (b)

A medium will not support an infinite number of standing waves of continuously different wavelengths

  1. True

  2. False


Correct Option: A
Explanation:

Only certain sized waves will stand on any one medium and thus a medium can be tuned to accept only certain waves or certain vibrations

Find the number of beats produced per sec by the vibrations $x _1=A\sin (320\pi t)$ and $x _2=A\sin (326\pi t)$.

  1. 3

  2. 4

  3. 5

  4. 6


Correct Option: A
Explanation:

$X _1=Asin(320\Pi t)$

$X _2=Asin(326\Pi t)$
On comparing it with general equation.
$X=Asin(wt)$
Then, $w _1=320\Pi $
$w _1=2\Pi f$
frequency=160 Hz
Similarly,
$w _2=326\Pi $
$w _2=2\Pi f$
frequency=163 Hz
No of beats=163-160=3

In an organ pipe(may be closed or open) of $99$ cm length standing wave is setup, whose equation is given by longitudinal displacement.
$\xi =(0.1mm)\cos \dfrac{2\pi}{0.8}(y+1cm)\cos 2\pi (400)t$
where y is measured from the top of the tube in meters and t is second. Here $1$cm is the end correction.
The air column is vibrating in :

  1. First overtone

  2. Fifth harmonic

  3. Third harmonic

  4. Fundamental mode


Correct Option: B

To and fro motion of a particle about its mean position is called :

  1. Frequency

  2. Amplitude

  3. Vibration

  4. Wavelength


Correct Option: C
Explanation:

two and fro motion of a particle about its mean position is called vibration 

so the answer is C.

The speed of mechanical waves depends on :-

  1. Density of medium

  2. Elasticity of medium

  3. Elasticity and density of medium

  4. Frequency of the wave.


Correct Option: C

A suspension bridge is to be built across valley where it is known that the wind can gust at $5\ s$ intervals. It is estimated that the speed of transverse waves along the span of the bridge would be $400\ m/s$. The danger of reasonant motions in the bridge at its fundamental frequency would be greater if the span had a length of :

  1. $2000\ m$

  2. $1000\ m$

  3. $400\ m$

  4. $80\ m$


Correct Option: B

A man generates a symmetrical pulse in a string by moving his hand up and down . At t = 0 the point in his hand moves downward. the pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position. then the point on the string at a distance 3m will reach its upper extreme first time at times t =

  1. $1.25 sec$

  2. $1 sec$

  3. $\frac{{13}}{{12}}\sec $

  4. $0.25$secs


Correct Option: D
Explanation:

The man's hand passes $6$ times from the mean position in $1$ sec, we can find that string creates $3$ cycles after $1$ sec.

Frequency of wave= $3Hz$
$V=f\lambda\Rightarrow \lambda=\cfrac {V}{f}=\cfrac {3}{3}=1m$
To reach upper extreme $\longrightarrow$ have to travel $3\lambda/4$ distance.
Time to travel $\cfrac {3 \lambda}{4}=\cfrac {3}{1}\times \cfrac {1}{3}=\cfrac {1}{4}=0.25$ sec

String 1 has twice the length, twice the radius, twice the tension and twice the density of another string 2. The relation between their fundamental frequencies of 1 and 2 is:

  1. $f _ { 1 } = 2 f _ { 2 }$

  2. $f _ { 1 } = 4 f _ { 2 }$

  3. $f _ { 2 } = 4 f _ { 1 }$

  4. $f _ { 1 } = f _ { 2 }$


Correct Option: B

In a reasonance tube experiment, a closed organ pipe of lenght $120$ cm is used. initially it is completely fiiled with water. It is vibrated with tuning fork of frequency $340$ Hz. To achieve reasonance the water level is lowered then (given ${V _{air}} = 340m/\sec $., neglect end correction):

  1. minimum lenght of water column to have the resonance is 45 cm.

  2. the distance between two successive nodes is 50 cm.

  3. the maximum lenght of water column to resonance is 95 cm.

  4. the distance between two successive nodes is 25 cm.


Correct Option: D

A string of length $1m$ and linear mass density $0.01kgm^{-1}$ is stretched to a tension of $100N$. When both ends of the string are fixed, the three lowest frequencies for standing wave are $f _{1}, f _{2}$ and $f _{3}$. When only one end of the string is fixed, the three lowest frequencies for standing wave are $n _{1}, n _{2}$ and $n _{3}$. Then 

  1. $n _{3} = 5n _{1} = f _{3} = 125 Hz $

  2. $f _{3} = 5f _{1} = n _{2} = 125 Hz $

  3. $f _{3} = n _{2} = 3f _{1} = 150 Hz $

  4. $n _{2} = \displaystyle \dfrac {f _{1} + f _{2}}{2} = 75 Hz $


Correct Option: D
Explanation:

When both ends are fixed, the string forms a length half the wavelength. That is, it has two nodes at the ends. For the next frequency, it will have the length equals the wavelength. So, the general formula for length of the string becomes $L = n \lambda /2$.


For the string fixed on only one end, there is always an anti node at one end and a node at the other end. So, the length of the string gets divided into $1/4th$ of the wavelength ($\lambda$). The general formula for the length of the string is $L' = n \lambda /4.$

The frequency $f$ becomes $V/ \lambda$, $V$ is the velocity. In the first case, frequency $f = nV/2L,$   $n = 1,2,3,....$

In the second case, it is $nV/4L$,    $n = 1,3,5,7......$ because of the length of the string will always have a half wave present. This makes n an odd number.
For the first case: 
$f _1 = 1/2L(\sqrt{(T/ \mu)}) = 50 Hz = V/2 \times L$
$f _2 = 2\times f _1 = 100 Hz = V/L$
$f3 = 3\times f _1 = 150 Hz = 3V/2\times L$

Second case:
$n _1 = V/4L$
$n _2 = 3V/4L = (f-1+f _2)/2 = (100+50)/2 = 75 Hz$

A massless rod of length $l$  is hung from the ceiling with the help of two identical wires attached at its ends. A block is hung on the rod at a distance $x$ from the left end. In the case, the frequency of the $1st$ harmonic of the wire on the left end is equal to the frequency of the $2nd$ harmonic of the wire on the right. The value of $x$ is

  1. $\displaystyle \dfrac{l}{2}$

  2. $\displaystyle \dfrac{l}{3}$

  3. $\displaystyle \dfrac{l}{4}$

  4. $\displaystyle \dfrac{l}{5}$


Correct Option: D
Explanation:

Since, the frequency of the first harmonic from the left is equal to that of second harmonic from right,
$ {\nu} _{1} = 2{\nu} _{2} $
Hence, $ {T} _{1} = {T} _{2} $
Thus, according to the question,
$ {T} _{1} (x) = {T} _{2} (l - x) $
Solving this equation for $ {T} _{1}$ and ${T} _{2} $ we get the value of x = $ \dfrac{l}{5} $

First overtone frequency of a closed pipe of length $l _1$ is equal to the$^{2nd}$ Harmonic frequency of an open pipe of length $l _2$. The ratio $l _1 \, l _2.$

  1. $3/4$

  2. $4/3$

  3. $3/2$

  4. $2/3$


Correct Option: C

The fundamental frequency of a stretched string is $V _o$. If the length is reduced by $35$% and tension increased by $69$% the fundamental frequency will be

  1. $2\, V _o$

  2. $0.5$

  3. $2.6$

  4. $1.6$


Correct Option: A

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is 20.000Hz)

  1. 6

  2. 4

  3. 5

  4. 7


Correct Option: D

The frequency of A note is $4$ times that of B note. The energies of two notes are equal. The amplitude of B note as compared to that of A note will be:

  1. double

  2. equal

  3. four times

  4. eight times


Correct Option: A
Explanation:

$E$ is for energy, $A$ for amplitude and $f $ for frequency.

As per the problem $E _{A} = E _{B}$
Hence, $f _{A} \times A _{A}^{2} = f _{B} \times A _B^2$
 $4f _{B} \times A _{A}^{2} = f _{B} \times A _B^2$
Hence, $2A _A = A _B $

A string vibrates in 5 segment to a frequency of 480 Hz. The frequency that will cause it to vibrate in 2 segments will be

  1. 96 Hz

  2. 192 Hz

  3. 1200 Hz

  4. 2400 Hz


Correct Option: B
Explanation:

5 segments implies $\lambda = \dfrac{2}{5}l$
$\nu = \dfrac{v}{\lambda} = \dfrac{5v}{2l} = 480Hz$
If the string is in 2 segments.
$\lambda = l$
$\nu = \dfrac{v}{\lambda} = \dfrac{2}{5} \dfrac{5v}{2l} = \dfrac{2}{5} 480 = 192Hz$
Hence option B is correct.

The vibrating body while playing a violin is ___________.

  1. wire

  2. the box of the violin

  3. both wire and box

  4. only air


Correct Option: C

If you set up the seven overtone on a string fixed at both ends, how many nodes and antinodes are set up in it?

  1. $6, 5$

  2. $5, 4$

  3. $4, 3$

  4. $3, 2$


Correct Option: C
Explanation:

Third mode of vibration or second overtone has three loops.
Its consist of $4$ nodes and $3$ antinodes.

A pipe of length $l _1$ closed at one end is kept in a chamber of gas density $1$. A second pipe open at both ends is placed in the second chamber of gas density $2$. The compressibility of both the gases is equal.Calculate the length of the second pipe if the frequency of the first overtone in both the cases is equal.

  1. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$

  2. $\displaystyle \dfrac{4}{3}l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

  3. $l _{1}\sqrt{\dfrac{\mathrm{p} _{2}}{\mathrm{p} _{1}}}$

  4. $l _{1}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$


Correct Option: B
Explanation:

$l _{1}=\displaystyle \dfrac{3}{4}\dfrac{\mathrm{v} _{1}}{\mathrm{f} _{1}}$ , $l _{2}=\displaystyle \dfrac{\mathrm{v} _{2}}{\mathrm{f} _{2}}$

$\dfrac{3\mathrm{v} _{1}}{4l _{1}}=\dfrac{\mathrm{v} _{2}}{l _{2}}$

$l _{2}=\displaystyle \dfrac{4l _{1}\mathrm{v} _{2}}{3\mathrm{v} _{1}}=\dfrac{4l _{1}}{3}\sqrt{\dfrac{\mathrm{p} _{1}}{\mathrm{p} _{2}}}$

A steel wire of mass $4.0\ g$ and length $80\ cm$ is fixed at the two ends. The tension in the wire is $50\ N$. The wavelength of the fourth harmonic of the fundamental will be

  1. $80\ cm$

  2. $60\ cm$

  3. $40\ cm$

  4. $20\ cm$


Correct Option: C
Explanation:

$m=$ mass per unit length

$=\cfrac { 4\times { 10 }^{ -3 } }{ 80\times { 10 }^{ -2 } } =0.005\quad Kg/m$

given $T=50 N$

$L=80 cm=0.8m$

$\therefore v=\sqrt { \cfrac { T }{ m }  } =\sqrt { \cfrac { 50 }{ 0.005 }  } =100m/sec$


Fundamental frequency 

${ f } _{ 0 }=\cfrac { 1 }{ 2L } \sqrt { \cfrac { T }{ m }  } \\ =\cfrac { 1 }{ 2\times 0.8 } \sqrt { \cfrac { 50 }{ 0.005 }  } \\ =625\quad Hz$


$\therefore { f } _{ 4 }$ Frequency of fourth harmonic 

$4{ f } _{ 0 }=4\times 62.5=250\quad Hz$

As we know

${ v } _{ 4 }={ f } _{ 4 }{ \lambda  } _{ 4 }\\ \therefore { \lambda  } _{ 4 }=\cfrac { { v } _{ 4 } }{ { f } _{ 4 } } =\cfrac { 100 }{ 250 } \\ =0.4m\quad \\ =40 cm$

The wave-function for a certain standing wave on a string fixed at both ends is $y\left( x,t \right) =0.5\sin { \left( 0.025\pi x \right)  } \cos { 500\ t } $ where $x$ and $y$ are in centimeters and t is in seconds. The shortest possible length of the string is: 

  1. $126\ cm$

  2. $160\ cm$

  3. $40\ cm$

  4. $80\ cm$


Correct Option: C
Explanation:

For the shortest possible length, it should be allowing fundamental frequency resonance.

In fundamental frequency
$L=\dfrac { \lambda  }{ 2 } \quad \quad \quad \quad \left( \because K=\dfrac { 2\pi  }{ \lambda  }  \right) $
$=\dfrac { 2\pi /K }{ 2 } =\dfrac { \pi  }{ K } $
from $y=0.5\sin\left( 0.025\pi x \right) \cos\left( 500t \right) $
$K=0.025\pi \quad \quad \quad (on\quad comparing\quad with\quad y=A\sin\left( Kx \right) \cos\left( wt \right) )$
$\therefore \quad L=\dfrac { \pi  }{ 0.025\pi  } =\dfrac { 1000 }{ 25 } $
$\left[ L=40cm \right] $

Hence Option (C) is correct.

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. The length of the open pipe is

  1. 8 m

  2. 4 m

  3. 2 m

  4. 1 m


Correct Option: B
Explanation:

$l _0=$ length of organ pipe (open)

$l _c=$ length of close organ pipe
frequency of open pipe$=\cfrac{3V}{2l _0}$
frequency of close pipe $=\cfrac{3V}{2l _c}$
$\therefore \cfrac{3V}{2l _0}=\cfrac{3V}{4l _c}\2l _0=4l _c\2\times2m=4m$
B is the correct option.

A guitar string is $90 cm$ long and has a fundamental frequency of $124 Hz$. To produce a fundamental frequency of $186 Hz$, the guitar should be pressed at ?

  1. $60 cm$

  2. $30 cm$

  3. $20 cm$

  4. $ 10 cm$


Correct Option: A
Explanation:

Here, $L _1 = 90 cm, \upsilon _1 = 124 Hz, \upsilon _2 = 186 Hz, L _2=?$

According to the law of length, $\upsilon _2L _2 = \upsilon _1L _1$

$\therefore \upsilon _2= \dfrac{\upsilon _1L _1}{\upsilon _2} = \dfrac{124 \times 90}{186} = 60 cm$

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