Tag: sonometer and laws of transverse vibrations

Questions Related to sonometer and laws of transverse vibrations

The fundamental frequency of sonometer wire is 600Hz. When length wire is shorted by 25%, the frequency of ${ 1 }^{ st }$ overtone will be

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. 800 Hz

  2. 1200 HZ

  3. 1600 Hz

  4. 2000 Hz

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Correct Option: C

In Melde's experiment when longitudinal position is used 4 loops are formed on string under tension of 16 g-wt . Now the string is replaced by another string of same material but of diameter half of the previous diameter and length half that of the original strings . What should be the tension in the string to obtain 2 loops on the strings , when B position is used ? 

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. 16 g-wt

  2. 32 g-wt

  3. 8 g-wt

  4. 4 g-wt

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Correct Option: B

A string of length $36cm$ was in unison with a fork of frequency $256Hz$. It was in unison with another fork when the vibrating length was $48cm$, the tension being unaltered. The frequency of second fork is   

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. $212Hz$

  2. $320Hz$

  3. $384Hz$

  4. $192Hz$

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Correct Option: D
Explanation:

$f=\dfrac{v }{2L}$
$v =f(2\ L)$
$=256\times 2\times 36$
$=18432\ cm/s.$


wave velocty remains same
$f=\dfrac{v }{2L}$
$=\dfrac{18432}{2\times 48}$
$=192\ Hz.$

The total mass of a wire remains constant on stretching the length of wire to four times. It's frequency will become:

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. 4 times

  2. 1/2 times

  3. 8 times

  4. $\sqrt{2}$ times

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Correct Option: B
Explanation:

Frequency, $f=\dfrac{1}{2l}\sqrt{\dfrac{t}{\mu }}$


Length is made four times, but mass is same.

$\Rightarrow$ Mass per unit length is $\mu'=\dfrac{\mu }{4}$

$\Rightarrow f'=\dfrac{1}{2(4l)}\sqrt{\dfrac{t}{\frac{\mu }{4}}}$$=\dfrac{2}{2(4l)}\sqrt{\dfrac{t}{\mu }}$ 

$\Rightarrow \dfrac{f'}{f}=\dfrac{1}{2}$

$\Rightarrow f'=\dfrac{f}{2}$

The length and diameter of a metal wire is doubled. The fundamental frequency of vibration will change from '$n$' to (Tension being kept constant and material of both the wires is same)

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. $\dfrac { n }{ 4 } $

  2. $\dfrac { n }{ 8 } $

  3. $\dfrac { n }{ 12 } $

  4. $\dfrac { n }{ 16 } $

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Correct Option: A
Explanation:

Fundamental frequency of vibration $n = \dfrac{v}{2L} \sqrt{\dfrac{T}{\mu}}$ 

where $\mu$ is the mass per unit length of the wire i.e. $\mu = \dfrac{M}{L}$
Mass of the wire $M = \rho (\dfrac{4\pi}{3} R^3)$
$\implies$ $n = \dfrac{v}{2L} .\sqrt{\dfrac{TL}{\rho \dfrac{4\pi }{3} R^3}}$
$\implies$ $n \propto \dfrac{1}{R\sqrt{LR}}$      .....(1)
Given :  $L _2 = 2L$  $R _2 = 2R$
From equation (1), we get  $\dfrac{n _2}{n} = \dfrac{R \sqrt{RL}}{R _2 \sqrt{L _2 R _2}}$
Or  $\dfrac{n _2}{n} = \dfrac{R \sqrt{R L}}{(2R) \sqrt{(2L) (2R)}}   = \dfrac{1}{4}$
$\implies$  $n _2 = \dfrac{n}{4}$

The frequency of vibration of a sonometer wire is directly proportional to linear density of the wire:

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

  1. True

  2. False

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Correct Option: B
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt{(T/\mu)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

The tension in a piano wire is $10 N$. The tension in a piano wire to produce a node of double frequency is

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

  1. $20 N$

  2. $40 N$

  3. $10 N$

  4. $120 N$

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Correct Option: B
Explanation:

For frequency of oscillation of wire $n \propto \sqrt{T}$. 

Here $T$ is tension in the wire. In order to increase frequency twice, tension needs to be made $4$ times. 
So, new tension must be $4 \times 10 = 40 N$

A knife edge divides a sonometer wire in two parts which differ in length by 2 mm. The whole length of the wire is 1 meter. The two parts of the string when sounded together produce one beat per second. Then the frequency of the smaller and longer pans.in Hz,are

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

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5

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Correct Option: D

A sonometer wire of length $l _1$ vibrates with a frequency 250 Hz. If the length of wire is increased then 2 beats/s are heard. What is ratio of the lengths of the wire?

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

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57

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Correct Option: A
Explanation:

The frequency of sonometer wire is given by
$n=\dfrac{p}{2l}\sqrt{(\dfrac{T}{m})}$
or $n\propto \dfrac{1}{l}$     ...(i)
$\therefore \dfrac{n _1}{n _2}=\dfrac{l _2}{l _1}$
or $\dfrac{250}{250-2}=\dfrac{l _2}{l _1}$
or $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

The tension in the sonometer wire is decreased by 4% by loosening the screws. It fundamental frequency

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

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary

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Correct Option: C
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

The fractional change in frequency with change in tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)