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

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)

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

$5\ beats/second$ are heard when a tuning fork is sounded with sonometer wire under tension, when the length of the sonometer wire is either $0.95\ m$ or $1\ m$. The frequency of the fork will be:

  1. $195\ Hz$

  2. $150\ Hz$

  3. $300\ Hz$

  4. $251\ Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When length is 0.95m

$v _1=\frac{v}{2\times 0.95}=\frac{v}{1.9}$
When length is 1m
$v _2=\frac{v}{2\times 1}=\frac{v}{2}$
$ v _1-v=5\quad v-v _2=5$
$ v _1-v _2=10$
$ \frac{v}{1.9}-\frac{v}{2}=10$
$\frac{0.1v}{3.8}=10$
$v=380m/s$
So, $v _1=200Hz , v _2=190Hz$
Then,
$v=195Hz$


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

A stone is hung in air from a wire, which is stretched over a sonometer. The bridges of the sonometer are 40 cm apart when the wire is in unison with a tuning fork of frequency 256 Hz. When the stone is completely immersed in water, the length between the bridges is 22 cm for re-establishing unison. The specific gravity of material of stone is 

  1. $
    \sqrt {\dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}}
    $

  2. $
    \dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}
    $

  3. $
    \dfrac{{40}}
    {{40 - 22}}
    $

  4. $
    \sqrt {\dfrac{{40}}
    {{40 - 22}}}
    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency f is proportional to sqrt(T)/L. In air, f = k*sqrt(Mg)/L1. In water, f = k*sqrt((M-m)g)/L2. Equating the two gives the ratio of weights, which leads to the specific gravity formula D/(D-d) = (L1/L2)^2.