Tag: vibrations of stretched strings

Questions Related to vibrations of stretched strings

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 

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

  1. 1 : 4 : 9 : 16

  2. 1 : 2 : 3 : 4

  3. 16 : 9 : 4 : 1

  4. 4 : 3 : 2 : 1

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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.

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

  1. 0.3 h

  2. 0.8 h

  3. 0.68 h

  4. 0.5 h

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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

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

  1. 1:2:3:4

  2. 16:9:4:1

  3. 1:4:9:16

  4. 4:3:2:1

Show answer
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

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

  1. vibrates with frequency of 416 Hz

  2. vibrates with frequency of 208 Hz

  3. vibrates with frequency of 832 Hz

  4. stop vibrating

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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

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

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

Show answer
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 

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

  1. 6 N

  2. 5 N

  3. 4 N

  4. 2 N

Show answer
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

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

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

Show answer
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 

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

  1. 3.6 kg decreased

  2. 3.6 kg increased

  3. 6.4 kg decreased

  4. 6.4 kg increased

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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

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

  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} $

Show answer
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 $3.00 kg$ ?

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

  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.

Show answer
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.