Tag: standing waves in strings

Questions Related to standing waves in strings

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

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. 3

  2. 4

  3. 5

  4. 6

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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 :

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. First overtone

  2. Fifth harmonic

  3. Third harmonic

  4. Fundamental mode

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Correct Option: B

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

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. Frequency

  2. Amplitude

  3. Vibration

  4. Wavelength

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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 :-

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. Density of medium

  2. Elasticity of medium

  3. Elasticity and density of medium

  4. Frequency of the wave.

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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 :

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. $2000\ m$

  2. $1000\ m$

  3. $400\ m$

  4. $80\ m$

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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 =

physics oscillations and waves standing waves in strings standing waves reflection of waves

  1. $1.25 sec$

  2. $1 sec$

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

  4. $0.25$secs

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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:

physics oscillations and waves standing waves in strings standing waves reflection of waves

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

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

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

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

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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):

physics oscillations and waves standing waves in strings standing waves reflection of waves

  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.

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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 

physics oscillations and waves standing waves in strings standing waves reflection of waves

  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 $

Show answer
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

physics oscillations and waves standing waves in strings standing waves reflection of waves

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

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

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

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

Show answer
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} $