Tag: sonometer and laws of transverse vibrations

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

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 $1:2:3$ ?

  1. $30 \ cm$

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

  3. $80\ cm$

  4. $40\ cm, 80\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$f _{1}:f _{2}:f _{3}=1:2:3$
$\Rightarrow \dfrac{1}{l _{1}}:\dfrac{1}{l _{2}}:\dfrac{1}{l _{3}}=1:2:3$
$\Rightarrow l _{1}:l _{2}:l _{3}=6:3:2$
$l _{1}=\left ( \dfrac{6}{2+3+6} \right )\times 110=60\ cm$
$l _{2}=\dfrac{3}{11}\times 110=30\ cm$
$l _{3}=\dfrac{2}{11}\times 110=20\ cm$

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

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 

  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. ${\dfrac{1}{n}}=\displaystyle \dfrac{1}{n _{1}}+\dfrac{1}{n _{2}}+\dfrac{1}{n _{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total length of string is $l=l _{1}+l _{2}+l _{3}$
but $f\propto \dfrac{1}{l}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}+\dfrac{1}{f _{3}}$

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

To increase the frequency by $20\%$, the tension in the string vibrating on a Sonometer has to be increased by

  1. $44\%$

  2. $33\%$

  3. $22\%$

  4. $11\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $44\%$

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

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 $= 8$)

  1. $\sqrt{8}$

  2. $\sqrt{6}$

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

  4. $\displaystyle \dfrac{1}{\sqrt{8}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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