Tag: melde's experiment

Quizzes (0) Questions (15)

Questions Related to melde's experiment

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  $\dfrac{n _2}{n} = \dfrac{R \sqrt{R L}}{(2R) \sqrt{(2L) (2R)}}   = \dfrac{1}{4}$
$\implies$  $n _2 = \dfrac{n}{4}$