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

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 :

  1. $\dfrac{n}{2\sqrt{2}}$
  2. $\sqrt{2}n$
  3. $\dfrac{n}{4}$
  4. $\dfrac{2n}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\dfrac{n}{4}$.
Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

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:

  1. 10 Hz

  2. 30 Hz

  3. 50 Hz

  4. 70 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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?

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
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).

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

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

  1. 100/101

  2. 99/100

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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

  1. increase due to buoyancy

  2. decrease

  3. remains unchanged

  4. increase due to viscosity of oil
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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:

  1. 30 N

  2. 9 N

  3. 90 N

  4. 10 N
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

A sonometer wire of length 114 cm is fixed at the both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio 1:3:4?

  1. at 36 cm and 84 cm from one end

  2. at 24 cmand 72 cm from one end

  3. at 48 cm and 96 cm from one end

  4. at 72 cm and 96 cm from one end
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequencies are in ratio 1:3:4, so lengths must be in ratio 1/1 : 1/3 : 1/4, which is 12:4:3. Total parts = 19. Lengths are (12/19)*114 = 72cm, (4/19)*114 = 24cm, and (3/19)*114 = 18cm. Placing bridges at 72cm and 96cm (72+24) creates segments of 72, 24, and 18 cm.

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

If the length of the wire of a sonometer is halved the value of resonant frequency will get:

  1. doubled

  2. halved

  3. four times

  4. eight times
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ f=\dfrac{1}{2e}\sqrt{\dfrac{t}{\mu }}$
$f\alpha \dfrac{1}{l}$
$\therefore \dfrac{f _{1}}{f _{2}}=\dfrac{l _{2}}{l _{2}}=\dfrac{1}{2}$
$\Rightarrow f _{2} = 2f _{1}$
$\therefore $ frequency is doubled 

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

A string vibrates in n loops, when the linear mass density is w gm/cm. If the string should vibrate in (n+2) loops, the new wire should have linear mass density:

  1. less than w

  2. more than w

  3. equal to w

  4. equal to w/2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

We also know that frequency is proportional to 1/ number of loops (n)

Thus, $n \alpha \sqrt(\mu)$

Larger n, larger should be the value of $\mu$

The correct option is (b)