Laws of vibrations of stretched strings - class-XI

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 $\sqrt(\mu)$

For frequency of oscillation of wire $n \propto \sqrt{T}$. 

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.

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 $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

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)

second overtone = third harmonic.
Thus, the length of the wire is $L=3\lambda/2$

The correct option is (c)

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

Thus, frequency of vibration is directly proportional to $\sqrt(T)$

$We\quad have\quad f=\frac { 1 }{ 2l } \sqrt { \frac { T }{ m }  } \ \ \therefore f\propto \frac { 1 }{ l } \quad \quad and\quad f\propto \frac { 1 }{ \sqrt { m }  } \ \ And\quad we\quad know\quad that\quad as\quad l\quad becomes\quad double\quad mass/unit\quad length\quad becomes\quad half.\ \ \therefore f\quad becomes\quad \frac { 1 }{ \sqrt { 2 }  } times\quad the\quad orignal.\ \ Hence\quad { f }^{ ' }=0.7f$

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)

$n\propto V$
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).

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)

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

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

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

The correct option is (c)

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.

$ 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 

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)

When length is 0.95m

Frequency f is proportional to sqrt(T)/L. In air, f = ksqrt(Mg)/L1. In water, f = ksqrt((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.

Fundamental frequency $\nu \propto \sqrt{T}$.
So, $\dfrac{\nu}{\nu'}=\sqrt{\dfrac{T}{T'}}\Rightarrow \dfrac{T}{T'}=\left(\dfrac{\nu}{\nu'}\right)^2=\dfrac{1}{4}$ 
$\Rightarrow T'=4T$. 

Fundamental frequency f = (1/2L) * sqrt(T/m). If f is constant and L is doubled, sqrt(T) must double, meaning T must increase by a factor of 4. Since T is proportional to the load, the load must be 4 * 4kg = 16kg.

The speed of a transverse wave in a stretched wire is given by v = sqrt(T / mu), where T is stress multiplied by cross-sectional area and mu is mass per unit length (density multiplied by area). Thus, v simplifies to sqrt(stress / density). Plugging in 3.0 x 10^8 and 7.5 x 10^5 gives sqrt(400) = 20 m/s.

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
if $L'=4L$
$f'=\dfrac{1}{8L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4}\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}=\dfrac{1}{4} f$
Option "A" is correct.

$\quad f \propto \sqrt T$
$f _{new} = \sqrt{0.81T} = 0.9\sqrt{T}$, that is
 or decrease of $ 10\mbox{%}$

$V=\sqrt { \dfrac { TL }{ m }  }$

$T:$tension

$m:$string mass

$L:$string length

$f=\dfrac { V }{ 2L } $          ----- fundamental frequency

suppose initial mass hanging is $M.$

$T=Mg$

${ f } _{ 1 }=\dfrac { V }{ 2L } $

${ f } _{ 1 }=\dfrac { 1 }{ 2L } \sqrt { \dfrac { MgL }{ m }  } $

$final \ \  mass=(M+8)$

${ f } _{ 2 }=\dfrac { 1 }{ 2L } $$\sqrt { \dfrac { (M+8)gL }{ m }  } $

 ${ f } _{ 2 }={ 3f } _{ 1 }$

On solving, we get

$M=1$ kg

the frequency at end and at the begining will remain same.
if one wave is generated at bottom in one sec, only one will reach the top.
hence we need equate frequency at top and bottom
$ \dfrac {{v} _{1}}{{\lambda} _{1}} = \dfrac {{v} _{2}}{{\lambda} _{2}} $
$ {v} _{2}    \alpha   \sqrt{tension   of   rope }    \alpha    \sqrt{m + M} $
$ {v} _{1}    \alpha   \sqrt{tension   of   rope }    \alpha    \sqrt{m } $

$f=\dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$
i.e. , for same sonometer, for different segments of bridges length will be the tuning parameter.
 we are given, 
$f _1:f _2:f _3=1:2:3$
while,
$f _1:f _2:f _3=\dfrac{1}{L _1}:\dfrac{1}{L _2}:\dfrac{1}{L _3}$
therefore $L _1:L _2:L _3=\dfrac{1}{f _1}:\dfrac{1}{f _2}:\dfrac{1}{f _3}=6:3:2$

Frequency $\nu \propto \dfrac{1}{l}$. So, $\dfrac{\nu}{\nu'}=\dfrac{l'}{l}$
$\Rightarrow 1-\dfrac{l'}{l}=1-\dfrac{\nu}{\nu'}=1-\dfrac{4}{5}=0.2$
$\Rightarrow \Delta l=0.2\times l =0.16m$ 

Fundamental frequency on the wire, $\nu = \dfrac{1}{2  L} \sqrt{\dfrac{T}{\mu}}$   

In the third harmonic, the wire has three segments separated by nodes, with maximum amplitude h at the antinodes. Point A is at L / 18, which corresponds to the middle of the first loop, meaning its amplitude of vibration is equal to the antinodal amplitude h multiplied by sin(3 * pi * (L/18) / L), which simplifies to h * sin(pi/6) = 0.5h.

Frequency f is proportional to sqrt(T). If tensions are in ratio 1:4:9:16, frequencies are in ratio sqrt(1):sqrt(4):sqrt(9):sqrt(16), which is 1:2:3:4.

since the wire is being excited by the tunning fork of frequency $ 416 $ Hz then wire of sonometer will always vibrate at frequency $ 416 $Hz the change in length will only effect its fundamental frequency. 

The correct answer is B

Let initial tension = T.
Then given that:
$2\nu = \sqrt{\frac{T}{\mu}}$ $2\nu$ is the frequency
Upon increasing the tension by 2.5N, the frequency becomes $3\nu$
$3\nu = \sqrt{\frac{T+2.5}{\mu}}$
Dividing the two equations.
$\frac{3}{2} =\sqrt{ \frac{T+2.5}{T}}$
$\frac{9}{4} = \frac{T+2.5}{T}$
$9T = 4T + 10$
$5T = 10$
$T = 2N$
Option d is correct.

Displacement y = A sin(alphax + betat + pi/6). At x=1, t=1: y = 7.5 * sin(0.56 + 12 + 0.52) = 7.5 * sin(13.08). Velocity v = dy/dt = Abetacos(alphax + betat + pi/6). Calculating these values yields approximately 3.75cm and 77.94 cm/s.

frequency in a sonometer is given as $f= \dfrac{v}{2l}\sqrt{\dfrac{T}{\mu}}$

$y={ Y } _{ 0 }\sin { 2\pi \left( ft-\frac { x }{ \lambda  }  \right)  } $

Since the mass becomes twice, the tension in the wire becomes twice.

$v=\sqrt{\dfrac{T}{mass\ per\ unit\ length}}=100\ m/s$
when tension is double $ T$ becomes $2T$ and mass per unit length remains same
so, $v _1 = \sqrt{\dfrac{2T}{mass\ per\ unit\ length}}$
$=\sqrt{2}\times\sqrt{\dfrac T{mass\ per\ unit\ length}}$
$=\sqrt{2}\times 100$
$= 141.4 m/s$

We know that, $\displaystyle n=\frac {1}{2} \sqrt {\frac {T}{m}} \Rightarrow n \alpha \sqrt {T}$
If tension is increased four times, the frequency will become twice.

$f \propto \sqrt{T}=\sqrt{mg}$
$\therefore m _{1}g _{1}=m _{2}g _{2}$
$\Rightarrow (1)g=m\left ( \dfrac{g}{6} \right )$
$\Rightarrow m=6kg$

$frequency\propto \dfrac{1}{no\ of\ loops}$
$f _{1}:f _{2}=6:4=3:2$
$f\alpha \sqrt{T}$
$\therefore \dfrac{3}{2}=\sqrt{\dfrac{65}{x}}$
$\dfrac{9}{4}=\dfrac{65}{x}$
$x=\dfrac{260}{9}=29\ gms$
$\therefore$ weight to be removed is $36 gms.$

Frequency f is proportional to sqrt(T). If T increases by a factor of 4, f increases by a factor of sqrt(4) = 2.

$f _{1}:f _{2}:f _{3}=1:2:3$
$\Rightarrow \dfrac{1}{l _{1}}:\dfrac{1}{l _{2}}:\dfrac{1}{l _{3}}=1:2:3$
$\Rightarrow l _{1}:l _{2}:l _{3}=6:3:2$
$l _{1}=\left ( \dfrac{6}{2+3+6} \right )\times 110=60\ cm$
$l _{2}=\dfrac{3}{11}\times 110=30\ cm$
$l _{3}=\dfrac{2}{11}\times 110=20\ cm$

Total length of string is $l=l _{1}+l _{2}+l _{3}$
but $f\propto \dfrac{1}{l}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{f _{1}}+\dfrac{1}{f _{2}}+\dfrac{1}{f _{3}}$

frequency increased by $20%
$$\Rightarrow f^{'}=\dfrac{6}{5}f$
$\therefore \sqrt{T^{'}}=\dfrac{6}{5}\sqrt{T}$
$\sqrt{T^{'}}=\sqrt{\dfrac{144}{100}}T$
$\therefore$ Tension is to be increased by $44%$

$f\propto \dfrac{1}{l}\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow l\propto \sqrt{T}$
$\therefore \dfrac{l _{1}}{l _{2}}=\sqrt{\dfrac{T _{1}}{T _{2}}}=\sqrt{\dfrac{8}{7}}$
$\dfrac{1}{\sqrt{7}l _{2}}=\dfrac{\sqrt{8}}{\sqrt{7}}$
$\Rightarrow l _{2}=\dfrac{1}{\sqrt{8}}$