Tag: laws of vibrations of stretched strings

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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 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

  1. $100 ms^{-1}$

  2. $141.4 ms^{-1}$

  3. $200 ms^{-1}$

  4. $282.8 ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $v _1 = \sqrt{\dfrac{2T}{mass\ per\ unit\ length}}$
$=\sqrt{2}\times\sqrt{\dfrac T{mass\ per\ unit\ length}}$
$=\sqrt{2}\times 100$
$= 141.4 m/s$

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

If the vibrations of a string are to be increased by a factor of two, then tension in the string should be made

  1. Twice

  2. Four times

  3. Eight times

  4. Half
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $1 kg$

  2. $\sqrt{6}$ kg

  3. $6 kg$

  4. $36 kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f \propto \sqrt{T}=\sqrt{mg}$
$\therefore m _{1}g _{1}=m _{2}g _{2}$
$\Rightarrow (1)g=m\left ( \dfrac{g}{6} \right )$
$\Rightarrow m=6kg$

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

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 :

  1. $72 \ gm$

  2. $36 \ gm$

  3. $21 \ gm$

  4. $29 \ gm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $36 gms.$

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

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:

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

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

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

  4. 2 m $s^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,$v=\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow v=\sqrt{\dfrac{25}{1/4}}=10m/s$
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

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