Tag: sonometer and laws of transverse vibrations

Questions Related to sonometer and laws of transverse vibrations

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

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

  1. 4

  2. (1/4)

  3. 2

  4. (1/2)

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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 $1:2:3$ ?

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

  1. $30 \ cm$

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

  3. $80\ cm$

  4. $40\ cm, 80\ cm$

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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 

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

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

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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

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

  1. $44\%$

  2. $33\%$

  3. $22\%$

  4. $11\%$

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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 $= 8$)

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

  1. $\sqrt{8}$

  2. $\sqrt{6}$

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

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

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Correct Option: D
Explanation:

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