Standing waves in strings - class-XI

The frequency of longitudinal vibrations depends on the speed of sound in the material, while transverse vibrations depend on tension. Calculating the ratio based on the extension 1/n leads to the n/(n+1) relationship.

Sounds are made by vibrations. Some vibrations are easy to see. The vibrations that create sound must travel through a medium, such as air or water, or anything made of molecules. With each forward motion, air molecules pulse outward, pushing other air molecules and crowding them together. With each backward motion, the molecules get less crowded. The forward and backward vibration of the glass creates a chain reaction of crowded and not-so-crowded molecules that ripples through the air. This traveling vibration is called a sound wave. 
Motion that moves to and fro in regular time intervals is called vibratory or oscillatory motion.

Answer is D.

When we hear a sound, we can identify its source from overtones present in the sound.
The fundamental is the frequency at which the entire wave vibrates. Overtones are other sinusoidal components present at frequencies above the fundamental. All of the frequency components that make up the total waveform, including the fundamental and the overtones, are called partials. Together they form the harmonic series.
Overtones which are perfect integer multiples of the fundamental are called harmonics. When an overtone is near to being harmonic, but not exact, it is sometimes called a harmonic partial, although they are often referred to simply as harmonics. Sometimes overtones are created that are not anywhere near a harmonic, and are just called partials or inharmonic overtones.

Vibrations that occur naturally after an initial disturbance, without continuous external driving force, are called free or natural vibrations.

Collect the data from the problem and find the length ($l$), tension ($T$) and mass ($M$) of the stretched string (b). 

All overtones are not stationary waves. Only those overtones which match the frequencies of the harmonics act as stationary waves.

All harmonics in a stringed instrument are stationary or standing waves

The correct option is (a)

$n^{th}$ overtone is always equal to (n+1) harmonic.

In a string fixed at one end, only odd harmonics are allowed

In a string fixed at one end, only odd harmonics are allowed and even harmonics are suppressed.

Thus the 10Hz standing wave is suppressed,
The correct option is (b)

The  third overtone is the 5th harmonic, since odd harmonics are only allowed in  a string fixed at one end

The correct option is (b)

Only certain sized waves will stand on any one medium and thus a medium can be tuned to accept only certain waves or certain vibrations

$X _1=Asin(320\Pi t)$

two and fro motion of a particle about its mean position is called vibration 

The speed of a mechanical wave in a medium is determined by the medium's elastic properties (which provide the restoring force) and its inertial properties (density). Both factors are essential for wave propagation.

Resonance occurs when the driving frequency matches the natural frequency of the system. For a bridge span, the fundamental frequency is f = v / (2L). Given v = 400 m/s and a period of 5s (frequency f = 0.2 Hz), setting 0.2 = 400 / (2L) leads to 0.4L = 400, so L = 1000 m.

The man's hand passes $6$ times from the mean position in $1$ sec, we can find that string creates $3$ cycles after $1$ sec.

The fundamental frequency of a stretched string is given by f = (1 / 2L) * sqrt(T / (pi * r^2 * d)). Substituting the doubled values for length L, radius r, tension T, and density d gives f1 = (1 / 2(2L)) * sqrt(2T / (pi * (2r)^2 * (2d))) = f2 / 4, which means f2 = 4 f1.

When both ends are fixed, the string forms a length half the wavelength. That is, it has two nodes at the ends. For the next frequency, it will have the length equals the wavelength. So, the general formula for length of the string becomes $L = n \lambda /2$.

Since, the frequency of the first harmonic from the left is equal to that of second harmonic from right,
$ {\nu} _{1} = 2{\nu} _{2} $
Hence, $ {T} _{1} = {T} _{2} $
Thus, according to the question,
$ {T} _{1} (x) = {T} _{2} (l - x) $
Solving this equation for $ {T} _{1}$ and ${T} _{2} $ we get the value of x = $ \dfrac{l}{5} $

Frequency f is proportional to (1 / L) * sqrt(T). If L is reduced by 35%, L_new = 0.65 L_old. If T is increased by 69%, T_new = 1.69 T_old. f_new / f_old = (L_old / L_new) * sqrt(T_new / T_old) = (1 / 0.65) * sqrt(1.69) = 1.3 / 0.65 = 2. Thus, f_new = 2 * f_old.

$E$ is for energy, $A$ for amplitude and $f $ for frequency.

5 segments implies $\lambda = \dfrac{2}{5}l$
$\nu = \dfrac{v}{\lambda} = \dfrac{5v}{2l} = 480Hz$
If the string is in 2 segments.
$\lambda = l$
$\nu = \dfrac{v}{\lambda} = \dfrac{2}{5} \dfrac{5v}{2l} = \dfrac{2}{5} 480 = 192Hz$
Hence option B is correct.

Third mode of vibration or second overtone has three loops.
Its consist of $4$ nodes and $3$ antinodes.

$m=$ mass per unit length

$=\cfrac { 4\times { 10 }^{ -3 } }{ 80\times { 10 }^{ -2 } } =0.005\quad Kg/m$

given $T=50 N$

$L=80 cm=0.8m$

$\therefore v=\sqrt { \cfrac { T }{ m }  } =\sqrt { \cfrac { 50 }{ 0.005 }  } =100m/sec$

Fundamental frequency 

${ f } _{ 0 }=\cfrac { 1 }{ 2L } \sqrt { \cfrac { T }{ m }  } \\ =\cfrac { 1 }{ 2\times 0.8 } \sqrt { \cfrac { 50 }{ 0.005 }  } \\ =625\quad Hz$

$\therefore { f } _{ 4 }$ Frequency of fourth harmonic 

$4{ f } _{ 0 }=4\times 62.5=250\quad Hz$

As we know

${ v } _{ 4 }={ f } _{ 4 }{ \lambda  } _{ 4 }\\ \therefore { \lambda  } _{ 4 }=\cfrac { { v } _{ 4 } }{ { f } _{ 4 } } =\cfrac { 100 }{ 250 } \\ =0.4m\quad \\ =40 cm$

For the shortest possible length, it should be allowing fundamental frequency resonance.

Here, $L _1 = 90 cm, \upsilon _1 = 124 Hz, \upsilon _2 = 186 Hz, L _2=?$

According to the law of length, $\upsilon _2L _2 = \upsilon _1L _1$

$\therefore \upsilon _2= \dfrac{\upsilon _1L _1}{\upsilon _2} = \dfrac{124 \times 90}{186} = 60 cm$