Tag: sonometer and laws of transverse vibrations

Questions Related to sonometer and laws of transverse vibrations

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. $L=3\lambda$

Show answer
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. $f$

Show answer
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. $\dfrac{2n}{\sqrt{2}}$

Show answer
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

Show answer
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)

Two identical sonometer wires have a fundamental frequency of $500$ Hz, when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of $5$ beats/sec, when both wires vibrate together?

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

  1. 2 

  2. 3

  3. 4

  4. 5

Show answer
Correct Option: A
Explanation:

$n\propto V$
and $V\propto \sqrt{T}$
$\Rightarrow n\propto \sqrt{T}$ ..$(1)$
$5$ beats/sec are obtained when the frequency of one become $505$ Hz i.e. percentage increase in frequency is $1\%$
From $(1)$ Percentage increase in $\eta =1\%$
$\Rightarrow \%$ increases in tension $=2\%$
(Note that method is applicable for small changes only).

When the length of the vibrating segment of a sonometer wire is increased by 1%, the percentage change in its frequency is

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

  1. 100/101

  2. 99/100

  3. 1

  4. 2

Show answer
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 length is given by $\Delta f/f = -\Delta L/L$.

Thus, the frequency decreases by 1%

The correct option is (c)

A brick is hung from a sonometer  wire. If the brick is immersed in oil, then frequency of the wire will 

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

  1. increase due to buoyancy

  2. decrease

  3. remains unchanged

  4. increase due to viscosity of oil

Show answer
Correct Option: B
Explanation:

When immersed in oil , the tension in the string decreases due to force of buoyancy.
$\therefore$ as $f\  \alpha\  \sqrt{T}$
frequency decreases  

The frequency of vibration of a sonometer doubles with doubling the length 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

Show answer
Correct Option: B
Explanation:

Frequency of vibration is inversely proportional to length. Hence doubling the length will make the frequency to be halved

The tension in a sonometer wire is found to be 90 N if the distance between the bridges is 30 cm. If the distance is reduced to 10 cm, the tension in the wire will be:

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

  1. 30 N

  2. 9 N

  3. 90 N

  4. 10 N

Show answer
Correct Option: C
Explanation:

The tension in the wire does not depend on the distance between the bridges and hence it remains constant

The correct option is (c)