Tag: sonometer and laws of transverse vibrations

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Questions Related to sonometer and laws of transverse vibrations

Multiple choice physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations 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

  1. 800 Hz

  2. 1200 HZ

  3. 1600 Hz

  4. 2000 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Fundamental frequency f = v / 2L. If length is shortened by 25%, L_new = 0.75L = 3/4 L. New fundamental frequency f_new = f / (3/4) = 4/3 * 600 = 800 Hz. The 1st overtone is the 2nd harmonic, which is 2 * f_new = 2 * 800 = 1600 Hz.

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

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 ? 

  1. 16 g-wt

  2. 32 g-wt

  3. 8 g-wt

  4. 4 g-wt
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency f = (p/L) * sqrt(T/m). Since f is constant, p1/L1 * sqrt(T1/m1) = p2/L2 * sqrt(T2/m2). m is proportional to diameter squared (d^2). Given d2 = d1/2, m2 = m1/4. Given L2 = L1/2. Substituting: 4/L1 * sqrt(16/m1) = 2/(L1/2) * sqrt(T2/(m1/4)). Solving for T2 gives 32 g-wt.

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

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   

  1. $212Hz$

  2. $320Hz$

  3. $384Hz$

  4. $192Hz$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


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

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

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

  1. 4 times

  2. 1/2 times

  3. 8 times

  4. $\sqrt{2}$ times
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }}$ 

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

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

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

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)

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

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

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

  4. $\dfrac { n }{ 16 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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  $\dfrac{n _2}{n} = \dfrac{R \sqrt{R L}}{(2R) \sqrt{(2L) (2R)}}   = \dfrac{1}{4}$
$\implies$  $n _2 = \dfrac{n}{4}$

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

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

  1. True

  2. False
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)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

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

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

  1. $20 N$

  2. $40 N$

  3. $10 N$

  4. $120 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $4 \times 10 = 40 N$

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

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

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the frequency formula f = (1/2L) * sqrt(T/m), the difference in frequencies for two segments of lengths L1 and L2 is 1 Hz. Given L1 + L2 = 1m and L1 - L2 = 0.002m, we find L1 = 0.501m and L2 = 0.499m. The frequencies are proportional to 1/L, leading to 124.5 Hz and 125.5 Hz.

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 $l _1$ 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?

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

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

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

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary
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 tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)