Wave velocity - class-XII

For all values of t, y$=0$ at $x=0$
Hence, (A) and (C) are correct.

The initial frequency is 205 Hz. When tightened, the string produces 6 beats in 2 seconds, meaning the beat frequency is 6 / 2 = 3 Hz. Tightening a string increases its tension and thus its frequency. Therefore, the new frequency must be 205 + 3 = 208 Hz.

For a wire vibrating in its fundamental mode, the frequency f = v / (2L). However, we can use the relation v = sqrt(T/mu). Given the mass M = 0.035 kg and linear density mu = 0.04 kg/m, the length L = M/mu = 0.875 m. The fundamental frequency f = v / (2L) = 45 Hz, so v = 45 * 2 * 0.875 = 78.75 m/s, which rounds to 79 m/s.

The transverse motion is given by y = A sin(omega * t + phi). Comparing y1 = 0.2 sin(3 * pi * t) with the standard form, omega = 3 * pi. Since omega = 2 * pi * f, we have 3 * pi = 2 * pi * f, which gives f = 1.5 Hz.

The fundamental frequency of a string is f = v / (2L). When divided into segments of lengths L1, L2, L3, the frequencies are f1 = v / (2L1) = n, f2 = v / (2L2) = 2n, f3 = v / (2L3) = 3n. The total length L = L1 + L2 + L3 = v/(2n) + v/(4n) + v/(6n) = (6+3+2)v / 12n = 11v / 12n. The fundamental frequency of the whole string is f0 = v / (2L) = v / (2 * 11v / 12n) = 6n / 11.

The frequency of a sonometer wire is inversely proportional to its length, f proportional to 1/L. Thus, f1 * L1 = f2 * L2, meaning f1 * 16.0 = f2 * 16.2. Also, the beat frequency is f2 - f1 = 5. Solving these simultaneous equations: 16.0 f1 = (f1 + 5) * 16.2, giving 16.0 f1 = 16.2 f1 + 81, so 0.2 f1 = 81, leading to f1 = 400 Hz and f2 = 405 Hz.

$\large \begin{array}{l} Here, \ y=20\sin  \frac { { \pi x } }{ 2 } \cos  40\pi t-----(i) \ compare, with, eqution,, (i),  \ \Rightarrow y=2r, , \sin  \frac { { 2\pi  } }{ \lambda  } , \times , , \cos  \frac { { 2\pi  } }{ \lambda  } Vt \ Now, \ \Rightarrow \frac { { 2\pi  } }{ \lambda  } =\frac { \pi  }{ 2 } , , and, , \frac { { 2\pi  } }{ \lambda  } V=40\pi  \ so, \ \Rightarrow \frac { \pi  }{ 2 } , \times V=40\pi  \ \therefore , , V=80, m/s \end{array}$

Equation of motion of harmonic oscillator is $y=Asin(\omega t+\phi _0)$

Answer is C.

A transverse wave is a wave in which the medium vibrates at right angles to the direction that the wave travels. An example of a transverse wave is a wave in a rope held with a hand on one end and tied to a pole on another end.
In this wave, energy is provided by a persons hand moving one end of the rope up and down. The direction of the wave is down the length of the rope away from the persons hand. The rope itself moves up and down as the wave passes through it.
The characteristic described in statement c is a property of all transverse waves, but not necessarily of all mechanical waves. A mechanical wave can also be longitudinal.
Hence, option C is correct and rest of the statements are incorrect.

The frequency of vibration of a string 

Given y = A sin(alpha x + beta t + pi/6). The displacement is y = 7.5 sin(0.56 * 1 + 12 * 1 + pi/6) = 7.5 sin(0.56 + 12 + 0.5236) = 7.5 sin(13.08) in radians, which calculates to approx 3.75 cm. The particle velocity is the time derivative of displacement, v = dy/dt = A beta cos(alpha x + beta t + pi/6) = 7.5 * 12 * cos(...) = 90 * cos(...) = 77.94 cm/s.

$\begin{array}{l} y=\dfrac { { 0.8 } }{ { 3{ x^{ 2 } }+12xt+12{ t^{ 2 } }+4 } }  \ =\dfrac { { 0.8 } }{ { 3\left( { { x^{ 2 } }+4x+4{ t^{ 2 } } } \right) +4 } }  \ =\dfrac { { 0.8 } }{ { 3{ { \left( { x+2t } \right)  }^{ 2 } }+4 } }  \ =\dfrac { { 0.8 } }{ { 3\times 4{ { \left( { \dfrac { x }{ 2 } +t } \right)  }^{ 2 } }+4 } }  \ \therefore Velocity=2m/s \end{array}$

Given frequency $f$=$\dfrac { 1 }{ t } $ and velocity $\nu$=10 m/s

Wave speed v = sqrt(T/mu) depends only on the properties of the medium (tension T and linear density mu). Particle velocity depends on the wave's amplitude and frequency, which are properties of the wave itself.

General equation of progressive wave:

$y = A \sin ( \omega t + k x )$

Here, $A$ is amplitude

$\omega$ is angular frequency

$k\,$is propagation constant

$t$  is time

$x$ is displacement

Given equation:

$y = 2 \sin ( 10 x + 300 t ).$

Linear charge density, $\mu =0.6\times {{10}^{-3}}\text{g/cm}\,\text{=}\,6\times \text{1}{{\text{0}}^{-3}}\,kg{{m}^{-1}}$
 

Compare with general equation:

Velocity of the wave in the string:

$v = \dfrac { \omega } { k }$$=\dfrac{300}{10}\text{=30}\,\text{m}{{\text{s}}^{-1}}$

Relation for velocity in terms of tension:

$T = \mu v ^ { 2 }$$=6\times {{10}^{-3}}\times {{(30)}^{2}}=5.4\,N$

Hence, tension in string is$5.4\,N$

For a string fixed at both ends, the standing wave must satisfy boundary conditions y = 0 at x = 0 and x = L. The form y = A sin(n * pi * x / L) * sin(omega * t) satisfies these conditions.

The equation of motion is y = A sin(pi t). The velocity of the particle is v = dy/dt = A pi cos(pi t). At t = 1/3 seconds, v = A pi cos(pi / 3) = A pi * (1/2) = (A pi) / 2.

The wave equation is y = A sin(omega * t - k * x). Here, omega = 100 and k = 1/10. Wave velocity v = omega / k = 100 / (1/10) = 1000 m/s.

The fundamental frequency of a stretched string is given by f = (1 / 2L) * sqrt(T / mu). Given L = 1 m, T = 20 N, and mu = 0.5 g/m = 0.5 * 10^-3 kg/m. Plugging in the values: f = (1 / 2(1)) * sqrt(20 / 0.5 * 10^-3) = 0.5 * sqrt(40000) = 0.5 * 200 = 100 Hz. The plucking point does not affect the fundamental frequency.

For SHM starting from rest at x = A, x(t) = A cos(omega * t). Distance traveled in time tau is A - A cos(omega * tau) = a. In next tau, distance is A cos(omega * tau) - A cos(2 * omega * tau) = 2a. Solving these equations leads to the amplitude being 4a.

$\begin{array}{l} w=4\pi  \ K=\dfrac { \pi  }{ { 16 } }  \ v=\dfrac { w }{ K } =64, m/s, along, , +x-axis \ Hence, \ option, , D, , is, correct, , answer. \end{array}$

For a standing wave y = A sin(kx) cos(omega t), the amplitude of a particle at position x is A_particle = |A sin(kx)|. A point midway between a node and an antinode corresponds to kx = pi/4. At this point, the amplitude is A * sin(pi/4) = A / sqrt(2). The frequency of vibration of any particle in a standing wave is the same as the wave frequency, which is omega / (2 pi).

When a block of mass m is suspended from an elastic wire of length l, area A, and Young's modulus Y, the effective spring constant of the wire is k = YA / l. The time period of vertical oscillations of the mass-spring system is given by T = 2 pi sqrt(m / k) = 2 pi sqrt(ml / YA).

$\begin{array}{l} For, , negative, , y-axis \ sign, , of, , \omega t, , & , , ky, , should, , be, , same,  \ x=A\sin  \left( { \omega t+ky } \right)  \ Hence, \ option, , B, , is, correct, , naswer. \end{array}$

Going by the data given to us, this wave is sinusoidal in nature and the wave equation takes the form of
$y = A\sin( \omega t - kx),$ as it is given that at $t = 0,$  the source is at mean position.
Here$,\ x = 4\ cm = 0.04\ m$
$y = A/2$
Amplitude $= A$
$t = \dfrac{T}{6}$
We know that $ \omega  = 2 \dfrac{ \pi }{T}$
$\Rightarrow \dfrac{A}{2} = A\sin((2  \pi  / T)(T/6) - 0.04k)$
$\sin((2 \pi  / T)(T/6) - 0.04k) = 1/2$
$\Rightarrow ((2  \pi / T)(T/6) - 0.04k) =  \pi / 6$
$k =  \pi  / 0.24$
wavelength $ \lambda = 2\pi  / k = 0.48\ m$

The total energy stored in one wavelength of a travelling wave on a string is given by E = (1/2) * mu * omega^2 * A^2 * lambda, where mu = F / v^2 and v = omega / k. Substituting these relations yields E = pi * k * F * A^2.

For a wave traveling in the positive x-direction, the function is f(t - x/v). Given f(t) = A sin(t/T), the wave equation is f(x, t) = A sin((t - x/v) / T) = A sin(t/T - x/(vT)).

$\begin{array}{l} \dfrac { { 3\lambda  } }{ 2 } =L \ \lambda =\dfrac { { 2l } }{ 3 }  \end{array}$

The wave speed v = sqrt(T/mu). At the bottom, T = m2 * g. At the top, T = (m1 + m2) * g. Since v = f * lambda and frequency f is constant, lambda is proportional to v. Thus, lambda_2 / lambda_1 = v_top / v_bottom = sqrt((m1 + m2) * g / (m2 * g)) = sqrt((m1 + m2) / m2).

$\begin{array}{l} v=\dfrac { 2 }{ { 2l } } \sqrt { \dfrac { { T\times l } }{ M }  }  \ or, , ,, , \sqrt { \dfrac { T }{ { Ml } }  } =\sqrt { \dfrac { { 20 } }{ { 5\times 10-4\times l } }  }  \ =\sqrt { 4\times { { 10 }^{ 4 } } } Hz=200Hz \end{array}$

The particle velocity is given by the partial derivative of y with respect to t. Since the wave is a standing wave, the velocity is v = dy/dt = -4 * 96 * pi * sin(pi*x/15) * sin(96 * pi * t). At t = 0.25 s, sin(96 * pi * 0.25) = sin(24 * pi) = 0, so the velocity is zero.

$ y=4\sin  \left( { 0.8x } \right) \cos  \left( { 400\pi t } \right)  \ w=400\pi =2\pi f \ f=200, Hz \ v=\dfrac { w }{ k } =\dfrac { { 400\pi \times 100 } }{ { 0.8 } } m/s \ 3\cdot \dfrac { v }{ { 2l } } =f \ \Rightarrow 200=\dfrac { { 3\times 400\pi \times 100 } }{ { 2\times l\times 0.8 } }  \ \Rightarrow l=\dfrac { { 3\times 400\pi  } }{ { 4\times 0.8 } }  \ =\dfrac { { 300\pi  } }{ { 0.8 } } , cm \ =11.8, m$

Beats are produced when two sound waves of slightly different frequencies interfere. When the two frequencies are identical, the beat frequency becomes zero, meaning the beat frequency vanishes, indicating that the string is properly tuned.

The tension at a distance h from the free end of a hanging rope is T = mu * g * h, where mu is the linear mass density. The wave speed is v = sqrt(T/mu) = sqrt(mu * g * h / mu) = sqrt(g * h).

We know, Speed of transverse wave $(v) = \sqrt{\dfrac{T}{\mu}}$

Linear mass density mu = mass / length = 2.5 kg / 20 m = 0.125 kg/m. Wave speed v = sqrt(T / mu) = sqrt(200 / 0.125) = sqrt(1600) = 40 m/s. Time taken t = distance / speed = 20 m / 40 m/s = 0.5 s.

The wave equation is y = 10 sin(0.01 * pi * x - 2 * pi * t). The transverse velocity is v_y = dy/dt = -10 * 2 * pi * cos(0.01 * pi * x - 2 * pi * t). The maximum transverse speed is 20 * pi = 20 * 3.14 = 62.8 cm/s.

${y _1} = 2\sin \left[ {4\pi t - 2\pi x} \right]$

$\begin{array}{l} y=7.5\sin  \left[ { 0.56\alpha +12t+\dfrac { \pi  }{ 6 }  } \right]  \ at, , x=1 \ & , , t=1 \ y=7.5\sin  \left[ { 0.56+12+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 12.56+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 4\pi +\dfrac { \pi  }{ 6 }  } \right]  \ =3.75, , cm \end{array}$

In a sine wave particles that are separated by a distance of odd multiple of half

From the given standing wave equation y = 5 sin(pi x / 3) cos(40 pi t), the wave number k = pi / 3. The separation between two consecutive nodes is equal to half of the wavelength, lambda / 2. Since lambda = 2 pi / k = 2 pi / (pi / 3) = 6 cm, the separation between consecutive nodes is 6 / 2 = 3 cm.

Given y = 4 sin(πx/15) cos(96π/t) - note this is TYPO: should be cos(96πt). Maximum displacement occurs when cos(96πt) = ±1. At x=5 cm, amplitude = 4 sin(π×5/15) = 4 sin(π/3) = 4 × √3/2 = 2√3 cm.

Tension at midpoint: The weight of the lower half of the string is (m/2) * g = 0.5 * 10 = 5 N. Linear mass density mu = 1 kg / 20 m = 0.05 kg/m. Wave speed v = sqrt(T / mu) = sqrt(5 / 0.05) = sqrt(100) = 10 m/s.

Hand completes 6 oscillations per second, so period T = 1/6 s. At t=0, hand moves downward. Pulse travels at 3 m/s. Point at 3 m reaches upper extreme when pulse arrives and phase is maximum. Time = distance/speed + T/4 = 3/3 + (1/6)/4 = 1 + 1/24 = 25/24 ≈ 0.25 s.

Tension at distance x from the free end is T = mu * g * x. Wave speed v = sqrt(T / mu) = sqrt(g * x). Thus, v is proportional to sqrt(x).

For a stretched string, T = Y * A * (delta_L / L). Wave speed v = sqrt(T / mu) = sqrt(Y * A * delta_L / (L * mu)). Since v is proportional to sqrt(delta_L), doubling the extension delta_L (from L/20 to L/10) increases the velocity by a factor of sqrt(2).

Average power P = 2 * pi^2 * f^2 * A^2 * mu * v. Tension T = 100 N, mu = 0.04 kg/m, so v = sqrt(100 / 0.04) = 50 m/s. P = 90 W, A = 0.05 m. 90 = 2 * 10 * f^2 * (0.05)^2 * 0.04 * 50. 90 = 20 * f^2 * 0.0025 * 2 = 0.1 * f^2. f^2 = 900, so f = 30 Hz.

The equation of stationary wave in a string is the amplitude of component waves is $4$mm.

$\begin{array}{l} a\sin  \left( { wt-kx } \right) +{ y _{ 1 } }=2a\sin  \cot  \cos  kx \ \Rightarrow y=a\sin  \left( { wt+kx } \right)  \ Ans., , (B) \end{array}$

$\begin{array}{l} Accorrding, , to, , question....................... \ Here, \ Difference, between, two, successive, crest, is, 2p. \ passes, difference, (\Delta \varphi )=\frac { { 2\pi  } }{ T }  \ , , , , , , , , , , , , , , , , , , , , , , , , , T=time, { { int } }erval(\Delta t) \ \therefore , , , n=2\pi =\frac { { 2\pi  } }{ T } \times 0.2 \ , , , , , , , , , , , , , , , \Rightarrow \frac { 1 }{ T } =5{ \sec ^{ -1 }  } \ \Rightarrow n=5Hz \ So, the, correct, option, is, B. \end{array}$

$\begin{array}{l} y=5\sin  \left( { \frac { { \pi x } }{ 3 }  } \right) .\cos  \left( { 4\pi t } \right)  \ Here, in, above, equation, , k=\frac { \pi  }{ 3 }  \ \frac { { 2\pi  } }{ \lambda  } =\frac { \pi  }{ 3 }  \ \lambda =6cm \ Hence,, dis\tan  ce, between, 2, nodes, is, \frac { \lambda  }{ 2 } =3cm \end{array}$

The equation is y = 4 sin(4 * pi * t - pi * x / 16). Comparing to y = A sin(omega * t - k * x), omega = 4 * pi and k = pi / 16. Wave speed v = omega / k = 4 * pi / (pi / 16) = 64 cm/s. Since the signs of omega * t and k * x are opposite, the wave travels in the +x direction.

The superposition of two waves traveling in opposite directions forms a standing wave. The distance between adjacent nodes is lambda / 2. Since k = 2 * pi / lambda, lambda / 2 = pi / k.

Wave speed v = 40 cm/s. At t = 2 s, y = 10 cos(x / 5). The general form is y = 10 cos(x / 5 - omega * t + phi). Since v = omega / k, omega = v * k = 40 * (1 / 5) = 8 rad/s. At t = 2, y = 10 cos(x / 5 - 16 + phi) = 10 cos(x / 5). Thus, phi = 16.

A wave pulse moving in the positive x-direction with speed c has the form f(x - ct). Substituting (x - ct) into the initial shape f(x) = a * e^(-x/b) gives a * e^(-(x - ct) / b).

The fundamental frequency of a stretched wire is inversely proportional to its length, f proportional to 1 / L, when tension and mass density remain constant. When the length is decreased from 1 m (100 cm) to 25 cm, the length is reduced by a factor of 4, so the fundamental frequency increases by a factor of 4. Thus, f2 = 256 * 4 = 1024 Hz.

In a stretched string, the restoring force is due to tension, which acts perpendicular to the displacement, supporting only transverse waves.

For a wave traveling in the negative x-direction, the argument is (omega * t + k * x). Given k = omega / c, the equation is y = a * cos(omega * t + omega * x / c).

A left travelling transverse wave given by $y=Asin(kx+\omega t)$

A wave traveling in +x direction is represented by $y=Asin(\omega t-kx)$

The waves with opposite phases superimpose to give a resultant wave of amplitude, $A-\dfrac{A}{3}=\dfrac{2A}{3}$ with phase angle $0$ at time $t=0$.

When two wave pulses travel on a string and meet, they obey the principle of superposition. They pass through each other without being permanently altered, maintaining their original shapes and velocities after the interaction.

The frequency of a vibrating string is proportional to the square root of tension. When immersed in water, the tension changes due to the buoyant force. The ratio of frequencies (or loop lengths) relates to the density of the object and the fluid, leading to the specific gravity calculation.

Speed of wave $(v) = 10 m/s$

Here, A = $5cm = 0.05$m, T = $0.2$s

The wave equation is y = 2 sin(pi * t - pi * x / v + pi / 3) where v = 2 cm/s. The slope is the partial derivative dy/dx. Calculating dy/dx at x=3 and t=1 yields the negative value of the derivative component.

frequency of fundamental note, $\upsilon = \dfrac{1}{2L} \sqrt{\dfrac{T}{m}}$

In first case, $256 = \dfrac{1}{2L} \sqrt{\dfrac{T}{m}} $                ......(i)

In second case, $320 =\dfrac{1}{2(L - 10)} \sqrt{\dfrac{T}{m}}$    ......(ii)

dividing (ii) and (i) we get,

$\dfrac{320}{256} = \dfrac{2L}{2(L - 10)}$ or $\dfrac{L}{L - 10} = \dfrac{5}{4}$

Given,

Firstly, we have to draw the pulse wave (see attached diagram).

If hand passes 6 times from the mean position in one second, then we know that string creates 3 wave lengths (λ) or 3 cycles after 1 second.

That means frequency (f) of the wave is $3 Hz.$

Now we can use below equation to calculate the value of wave length.

$V = f\lambda$ (V = velocity of the wave)

$\lambda =\dfrac{ V}{f}$

  $= \dfrac{(3m/s)}{3} = 1 m$

If $\lambda = 1m$, the point which have 3m distance is located at no.(6) (in the diagram).

to reach its upper extreme ----> have to travel 3λ/4 distance

time to travel $3\lambda = 1$ second

time to travel $\lambda = \dfrac{1}{3}$ seconds

time to travel $\dfrac{3\lambda}{4} = (\dfrac{1}{3}) \times (\dfrac{3}{4})$ seconds

 $= \dfrac{1}{4}$ seconds $= 0.25$ seconds

The average power of a wave is proportional to the square of the amplitude and the square of the frequency. If the frequency remains constant as the wave travels, the reduction in power must be due to a reduction in amplitude.

For a standing wave y = A sin(kx) cos(omegat), the nodes occur where sin(kx) = 0. For a rod fixed at both ends, the length L must be an integer multiple of half-wavelengths. Given k = 2pi / 40, lambda = 40. Minimum length is lambda / 2 = 20.

Velocity of wave in string A, ${v _A} = \sqrt {\dfrac{T}{{{\mu _A}}}}  = \sqrt {\frac{{200}}{2}}  = 10,,m/s$

When a wave reflects from a denser medium, it undergoes a phase change of 180 degrees. The amplitude of the reflected wave is determined by the energy reflection coefficient (R = sqrt(0.64) = 0.8).

Thicker, tighter strings, have a more "focussed" sound. They reach their resonant frequency more quickly, because the extra tension leaves them less scope to flap around.

Thicker, tighter strings, plucked the same distance, are louder, because they contain more energy. There is more kinetic energy to be transmitted to the sounding board.

As frequency $(f)$ is directly proportional to the square root of tension$(t)$.