Tag: vibrations of stretched strings

Questions Related to vibrations of stretched strings

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)

The sonometer wire is vibrating in the second overtone. The length of the wire in terms of wavelength is:

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

  1. $L=\lambda/2$

  2. $L=5\lambda/2$

  3. $L=3\lambda/2$

  4. $L=3\lambda$

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Correct Option: C
Explanation:

second overtone = third harmonic.
Thus, the length of the wire is $L=3\lambda/2$

The correct option is (c)

The frequency of vibration of a sonometer wire is inversely proportional to tension in the wire

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

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

Thus, frequency of vibration is directly proportional to $\sqrt(T)$

A wire has frequency f. Its length is doubled by stretching. Its frequency now will be:

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

  1. $1.4\ f$

  2. $0.7\ f$

  3. $2\ f$

  4. $f$

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Correct Option: B
Explanation:

$We\quad have\quad f=\frac { 1 }{ 2l } \sqrt { \frac { T }{ m }  } \ \ \therefore f\propto \frac { 1 }{ l } \quad \quad and\quad f\propto \frac { 1 }{ \sqrt { m }  } \ \ And\quad we\quad know\quad that\quad as\quad l\quad becomes\quad double\quad mass/unit\quad length\quad becomes\quad half.\ \ \therefore f\quad becomes\quad \frac { 1 }{ \sqrt { 2 }  } times\quad the\quad orignal.\ \ Hence\quad { f }^{ ' }=0.7f$

Fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubled keeping the tension same, then the new fundamental frequency is :

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

  1. $\dfrac{n}{2\sqrt{2}}$

  2. $\sqrt{2}n$

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

  4. $\dfrac{2n}{\sqrt{2}}$

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Correct Option: C
Explanation:
Frequency of vibration of sonometer wire is given as
$n=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}=\dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2d}}\Rightarrow n\propto \dfrac{1}{\sqrt{(d)}}$
If the length and diameter of the wire are doubled. The new frequency will be $\dfrac{n}{4}$.

A wire with linear density of 3 gm/mm is used as a sonometer wire for producing vibrations of frequency 50 Hz. This length of this wire is now halved, while the tension is reduced by 1/4th of the initial tension. What will be the frequency of vibrations produced:

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

  1. 10 Hz

  2. 30 Hz

  3. 50 Hz

  4. 70 Hz

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Correct Option: C
Explanation:

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

since length is halved and tension is made 1/4th the initial tension, their ratio $\sqrt(T)/L$ remains constant. Thus, frequency now depends only on linear mass density

Since the same wire is used, linear mass density dosen't change and hence the frequency also remains same

The correct option is (c)