Tag: sonometer and laws of transverse vibrations

Questions Related to sonometer and laws of transverse vibrations

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

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

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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} $

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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.

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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

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

  1. $100 ms^{-1}$

  2. $141.4 ms^{-1}$

  3. $200 ms^{-1}$

  4. $282.8 ms^{-1}$

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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

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

  1. Twice

  2. Four times

  3. Eight times

  4. Half

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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

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

  1. $1 kg$

  2. $\sqrt{6}$ kg

  3. $6 kg$

  4. $36 kg$

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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 :

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

  1. $72 \ gm$

  2. $36 \ gm$

  3. $21 \ gm$

  4. $29 \ gm$

Show answer
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:

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

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

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

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

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

Show answer
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,$v=\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow v=\sqrt{\dfrac{25}{1/4}}=10m/s$