Question 2

   Question: 2 The displacement of a travelling wave $y = C \sin [\frac{2\pi}{\lambda}(at - x)]$ where $t$ is time, $x$ is distance and $\lambda$ is the wavelength, all in S.I. units. Then the frequency of the wave is

   Options: 

    A. $2\pi\lambda/a$  

    B. $2\pi a/\lambda$  

    C. $\lambda/a$  

    D. $a/\lambda$

   Correct Answer: D

   Year: [NEET 2024 Re]

   Solution:  

    $y = c \sin \frac {2 \pi}{\lambda} (\mathrm {a t} - x)$  

    $y = c \sin \left(\frac {2 \pi}{\lambda} \mathrm {a t} - \frac {2 \pi}{\lambda} x\right)$  

    Comparing with $y = A \sin(\omega t - kx)$  

    $\omega = 2 \pi f = \frac {2 \pi a}{\lambda}$  

    $f = \frac {a}{\lambda}$

   Step Solution:

    1.  Expand the given wave equation: $y = C \sin \left( \frac{2\pi at}{\lambda} - \frac{2\pi x}{\lambda} \right)$.

    2.  Identify the standard form: Compare this with $y = A \sin(\omega t - kx)$.

    3.  Extract Angular Frequency ($\omega$): The coefficient of $t$ is $\omega$, so $\omega = \frac{2\pi a}{\lambda}$.

    4.  Relate $\omega$ to Frequency ($f$): Use the formula $\omega = 2\pi f$.

    5.  Calculate $f$: $2\pi f = \frac{2\pi a}{\lambda} \implies f = \frac{a}{\lambda}$.

   Difficulty Level: Easy

   Concept Name: Comparison of Traveling Wave Equations

   Shortcut Solution: The frequency $f$ is always the coefficient of time ($t$) divided by $2\pi$. In $y = C \sin [ \frac{2\pi a}{\lambda} t - \frac{2\pi}{\lambda} x ]$, the coefficient is $\frac{2\pi a}{\lambda}$. Thus, $f = (\frac{2\pi a}{\lambda}) / 2\pi = \frac{a}{\lambda}$.

Question 27

   Question: A wave travelling in the +ve x-direction having displacement along y-direction as 1 m, wavelength 2π m and frequency of $\frac { 1 } { \pi } \mathbf { H } \mathbf { z }$ is represented by

   Options: 

    A. $\mathbf { y } = \sin ( 1 0 \pi \mathbf { x } - 2 0 \pi \mathrm { t } )$  

    B. $\mathbf { y } = \sin ( 2 \pi \mathbf { x } + 2 \pi \mathbf { t } )$  

    C. $\mathrm { y } = \sin ( \mathrm { x } - 2 \mathrm { t } )$  

    D. $\mathbf { y } = \sin ( 2 \pi \mathbf { x } - 2 \pi \mathbf { t } )$

   Correct Answer: C

   Year: (2013 NEET)

   Solution: (Derived from parameters given in Question 27)

   Step Solution:

    1.  List given values: Amplitude $A = 1$ m, Wavelength $\lambda = 2\pi$ m, Frequency $f = \frac{1}{\pi}$ Hz.

    2.  Calculate Wave Number ($k$): $k = \frac{2\pi}{\lambda} = \frac{2\pi}{2\pi} = 1$ rad/m.

    3.  Calculate Angular Frequency ($\omega$): $\omega = 2\pi f = 2\pi \left(\frac{1}{\pi}\right) = 2$ rad/s.

    4.  Determine wave direction: For a wave in the +ve x-direction, the equation is $y = A \sin(kx - \omega t)$ or $y = A \sin(\omega t - kx)$.

    5.  Substitute values: $y = 1 \sin(1 \cdot x - 2 \cdot t) = \sin(x - 2t)$.

   Difficulty Level: Easy

   Concept Name: General Wave Equation Construction

   Shortcut Solution: For a wave moving in the +ve x-direction, the $x$ and $t$ terms must have opposite signs. Only options A, C, and D qualify. Calculate $k = 2\pi/\lambda = 1$. Only option C has $x$ with a coefficient of 1.

Question 33

   Question: The equation of a simple harmonic wave is given by $\mathbf { y } = 3 \mathbf { s } \mathbf { i n } \frac{\pi}{2} ( \bar { \mathbf { 50 t - x } } )$. where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is

   Options: 

    A. 2π  

    B. $\frac { 3 } { 2 } \pi$  

    C. 3π  

    D. $\frac { 2 } { 3 } \pi$

   Correct Answer: B

   Year: (2012 Mains)

   Solution:  

    $y = 3 \sin \left(2 5 \pi t - \frac {\pi}{2} x\right)$  

    $\omega = 2 5 \pi , k = \frac {\pi}{2}$  

    Wave velocity, $v = \frac{\omega}{k} = \frac{25\pi}{\pi/2} = 50 \text{ m/s}$  

    $v_p = \frac{dy}{dt} = 75 \pi \cos(25 \pi t - \frac{\pi}{2})$  

    $(v_p)_{max} = 75 \pi$  

    $\frac{(v_p)_{max}}{v} = \frac{75\pi}{50} = \frac{3}{2} \pi$

   Step Solution:

    1.  Expand the equation: $y = 3 \sin(25\pi t - \frac{\pi}{2}x)$.

    2.  Identify parameters: Amplitude $A = 3$, $\omega = 25\pi$, and $k = \frac{\pi}{2}$.

    3.  Calculate Wave Velocity ($v$): $v = \frac{\omega}{k} = \frac{25\pi}{\pi/2} = 50$ m/s.

    4.  Calculate Max Particle Velocity ($v_{p\_max}$): $v_{p\_max} = A\omega = 3 \times 25\pi = 75\pi$ m/s.

    5.  Find the ratio: $\text{Ratio} = \frac{75\pi}{50} = \frac{3\pi}{2}$.

   Difficulty Level: Medium

   Concept Name: Particle Velocity vs. Wave Velocity

   Shortcut Solution: The ratio of maximum particle velocity to wave velocity is always equal to $Ak$ (Amplitude $\times$ wave number). Here $A = 3$ and $k = \frac{\pi}{2}$. Therefore, Ratio $= 3 \times \frac{\pi}{2} = \frac{3\pi}{2}$.

Question 35

   Question: Two waves are represented by the equations $y_1 = a \sin(\omega t + kx + 0.57) \text{ m}$ and $y_2 = a \cos(\omega t + kx) \text{ m}$, where $x$ is in meter and $t$ in sec. The phase difference between them is

   Options: 

    A. 1.0 radian  

    B. 1.25 radian  

    C. 1.57 radian  

    D. 0.57 radian

   Correct Answer: A

   Year: (2011)

   Solution:  

    $y_1 = a \sin(\omega t + kx + 0.57) \therefore \text{phase } \phi_1 = \omega t + kx + 0.57$  

    $y_2 = a \cos(\omega t + kx) = a \sin\left(\omega t + kx + \frac{\pi}{2}\right)$  

    $\therefore \text{phase } \phi_2 = \omega t + kx + \frac{\pi}{2}$  

    $\text{Phase } \Delta \phi = \phi_2 - \phi_1 = \left(\omega t + kx + \frac{\pi}{2}\right) - (\omega t + kx + 0.57) = \frac{\pi}{2} - 0.57$  

    $= (1.57 - 0.57) \text{ radian} = 1 \text{ radian}$

   Step Solution:

    1.  Identify initial phase of first wave: $\phi_1 = \omega t + kx + 0.57$.

    2.  Convert cosine to sine for comparison: $a \cos(\theta) = a \sin(\theta + \frac{\pi}{2})$.

    3.  Identify phase of second wave: $\phi_2 = \omega t + kx + \frac{\pi}{2} = \omega t + kx + 1.57$.

    4.  Calculate the difference: $\Delta \phi = \phi_2 - \phi_1 = (\omega t + kx + 1.57) - (\omega t + kx + 0.57)$.

    5.  Final result: $\Delta \phi = 1.57 - 0.57 = 1.0 \text{ radian}$.

   Difficulty Level: Easy

   Concept Name: Phase difference between Sine and Cosine functions.

   Shortcut Solution: The phase difference between $\sin(\theta)$ and $\cos(\theta)$ is exactly $\pi/2$ (approx 1.57). Since the sine wave already has an added phase of 0.57, the net difference is $1.57 - 0.57 = 1.0$.

Question 38

   Question: A transverse wave is represented by $y = A \sin(\omega t - kx)$. For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

   Options: 

    A. $\pi A / 2$  

    B. $\pi A$  

    C. $2\pi A$  

    D. $A$

   Correct Answer: C

   Year: (2010)

   Solution:  

    The given wave equation is $y = A \sin(\omega t - kx)$  

    Wave velocity $v = \omega/k$  

    Particle velocity, $v_p = \frac{dy}{dt} = A \omega \cos(\omega t - kx)$  

    Maximum particle velocity, $(v_p)_{max} = A\omega$  

    According to the given question $v = (v_p)_{max}$  

    $\frac{\omega}{k} = A\omega \implies \frac{1}{k} = A \implies \frac{\lambda}{2\pi} = A \implies \lambda = 2\pi A$

   Step Solution:

    1.  Define Wave Velocity ($v$): $v = \frac{\omega}{k}$.

    2.  Find Maximum Particle Velocity ($v_{p\_max}$): $v_{p\_max} = A\omega$.

    3.  Set them equal as per condition: $\frac{\omega}{k} = A\omega$.

    4.  Simplify for Wave Number ($k$): $1/k = A$.

    5.  Relate $k$ to Wavelength ($\lambda$): Substitute $k = \frac{2\pi}{\lambda}$ into the equation to get $\frac{\lambda}{2\pi} = A$, hence $\lambda = 2\pi A$.

   Difficulty Level: Medium

   Concept Name: Relation between Wave Velocity and Particle Velocity.

   Shortcut Solution: Use the fixed ratio: $\text{Ratio} = \frac{v_{p\_max}}{v} = Ak$. For them to be equal, $Ak = 1$. Substituting $k = 2\pi/\lambda$ gives $A(2\pi/\lambda) = 1$, so $\lambda = 2\pi A$.

Question 42

   Question: A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x axis with a speed of 128 m/sec and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is

   Options: 

    A. $y = (0.02) \text{ m sin} (15.7x - 2010t)$  

    B. $y = (0.02) \text{ m sin} (15.7x + 2010t)$  

    C. $y = (0.02) \text{ m sin} (7.85x - 1005t)$  

    D. $y = (0.02) \text{ m sin} (7.85x + 1005t)$

   Correct Answer: C

   Year: (2009)

   Solution:  

    Amplitude $= 2 \text{ cm} = 0.02 \text{ m}$, $v = 128 \text{ m/s}$  

    $\lambda = 4/5 = 0.8 \text{ m}; \nu = \frac{128}{0.8} = 160 \text{ Hz}$  

    $\omega = 2\pi\nu = 2\pi \times 160 = 1005; k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.8} = 7.85$  

    $\therefore y = 0.02 \sin(7.85x - 1005t)$

   Step Solution:

    1.  Find Wavelength ($\lambda$): 5 waves in 4m means $\lambda = 4/5 = 0.8 \text{ m}$.

    2.  Calculate Wave Number ($k$): $k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.8} \approx 7.85 \text{ m}^{-1}$.

    3.  Calculate Frequency ($f$): $f = \frac{v}{\lambda} = \frac{128}{0.8} = 160 \text{ Hz}$.

    4.  Calculate Angular Frequency ($\omega$): $\omega = 2\pi f = 2 \times 3.14 \times 160 \approx 1005 \text{ rad/s}$.

    5.  Write equation for +ve x-direction: $y = A \sin(kx - \omega t) = 0.02 \sin(7.85x - 1005t)$.

   Difficulty Level: Hard

   Concept Name: General Wave Equation Parameters.

   Shortcut Solution: Since the wave moves in the positive x-direction, the signs between $x$ and $t$ terms must be opposite (ruling out B and D). Calculate $k = 2\pi/0.8 = 7.85$. Only option C matches this $k$ value.

Question 45

   Question: The wave described by $y = 0.25 \sin(10\pi x - 2\pi t)$, where $x$ and $y$ are in meters and $t$ in seconds, is a wave travelling along the

   Options: 

    A. +ve x direction with frequency 1 Hz and wavelength $\lambda = 0.2$ m  

    B. -ve x direction with amplitude $0.25$ m and wavelength $\lambda = 0.2$ m  

    C. -ve x direction with frequency 1 Hz.  

    D. +ve x direction with frequency $7t$ Hz and wavelength $\lambda = 0.2$ m

   Correct Answer: A

   Year: 2008

   Solution:  

    $y = 0.25 \sin(10\pi x - 2\pi t)$  

    $y_{max} = 0.25$  

    $k = \frac{2\pi}{\lambda} = 10\pi \implies \lambda = 0.2$ m  

    $\omega = 2\pi\nu = 2\pi \implies \nu = 1$ Hz  

    The sign is negative inside the bracket. Therefore this wave travels in the positive x-direction.

   Step Solution:

    1.  Identify Wave Parameters: From $y = A \sin(kx - \omega t)$, we have $k = 10\pi$ and $\omega = 2\pi$.

    2.  Determine Direction: Because the signs of the $x$ and $t$ terms are opposite (one positive, one negative), the wave travels in the positive x-direction.

    3.  Calculate Wavelength ($\lambda$): Using $\lambda = \frac{2\pi}{k} = \frac{2\pi}{10\pi} = \mathbf{0.2 \text{ m}}$.

    4.  Calculate Frequency ($\nu$): Using $\nu = \frac{\omega}{2\pi} = \frac{2\pi}{2\pi} = \mathbf{1 \text{ Hz}}$.

   Difficulty Level: Easy

   Concept Name: Standard Equation of Traveling Waves

   Shortcut Solution: Wave speed direction is determined by the sign between $kx$ and $\omega t$ (negative sign = positive direction). Frequency is always the coefficient of $t$ divided by $2\pi$ ($2\pi / 2\pi = 1$).

Question 49

   Question: A transverse wave propagating along x-axis is represented by $y(x, t) = 8.0 \sin(0.5\pi x - 4\pi t - \frac{\pi}{4})$ where $x$ is in metres and $t$ is in seconds. The speed of the wave is

   Options: 

    A. 8 m/s  

    B. 4π m/s  

    C. 0.5π m/s  

    D. π/4 m/s

   Correct Answer: A

   Year: 2006

   Solution:  

    $y(x, t) = 8.0 \sin(0.5\pi x - 4\pi t - \frac{\pi}{4})$  

    Compare with a standard wave equation, $y = a \sin(\frac{2\pi x}{\lambda} - \frac{2\pi t}{T} + \phi)$  

    we get $\frac{2\pi}{\lambda} = 0.5\pi$ or, $\lambda = \frac{2\pi}{0.5\pi} = 4$ m  

    $\frac{2\pi}{T} = 4\pi$ or, $T = \frac{2\pi}{4\pi} = \frac{1}{2}$ sec.  

    $\nu = \frac{1}{T} = 2$ Hz  

    Wave velocity, $v = \lambda \nu = 4 \times 2 = 8$ m/sec

   Step Solution:

    1.  Extract $k$ and $\omega$: From the equation, $k = 0.5\pi$ and $\omega = 4\pi$.

    2.  Identify Formula: Wave speed $v = \frac{\omega}{k}$.

    3.  Substitution: $v = \frac{4\pi}{0.5\pi}$.

    4.  Calculate: $v = \frac{4}{0.5} = \mathbf{8 \text{ m/s}}$.

   Difficulty Level: Easy

   Concept Name: Wave Speed from Wave Equation

   Shortcut Solution: Wave speed is simply the coefficient of $t$ divided by the coefficient of $x$. Here, $4\pi / 0.5\pi = 8$.

Question 52

   Question: The phase difference between two waves, represented by  

    $y_1 = 10^{-6} \sin[100t + (x/50) + 0.5]$ m  

    $y_2 = 10^{-6} \cos[100t + (x/50)]$ m  

    where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately.

   Options: 

    A. 1.07 radians  

    B. 2.07 radians  

    C. 0.5 radians  

    D. 1.5 radians

   Correct Answer: A

   Year: 2004

   Solution:  

    $y_1 = 10^{-6} \sin[100t + (x/50) + 0.5]$  

    $y_2 = 10^{-6} \cos[100t + (x/50)] = 10^{-6} \sin[100t + (x/50) + \frac{\pi}{2}]$  

    $= 10^{-6} \sin[100t + (x/50) + 1.57]$  

    The phase difference $= 1.57 - 0.5 = 1.07$ radians

   Step Solution:

    1.  Identify phase of $y_1$: $\phi_1 = 100t + (x/50) + 0.5$.

    2.  Convert $y_2$ to Sine: Use the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$.

    3.  Identify phase of $y_2$: $\phi_2 = 100t + (x/50) + 1.57$ (since $\frac{\pi}{2} \approx 1.57$).

    4.  Find Difference: $\Delta\phi = \phi_2 - \phi_1 = (100t + \frac{x}{50} + 1.57) - (100t + \frac{x}{50} + 0.5)$.

    5.  Calculate: $1.57 - 0.5 = \mathbf{1.07 \text{ radians}}$.

   Difficulty Level: Medium

   Concept Name: Phase Difference Comparison

   Shortcut Solution: A cosine wave leads a sine wave by $\pi/2$ (1.57 rad). Since the sine wave already has a phase of 0.5 rad, the net difference is $1.57 - 0.5 = 1.07$ rad.

Question 56

   Question: If a wave travelling in positive x-direction with $A = 0.2$ m, velocity $= 360$ m/s and $\lambda = 60$ m, then correct expression for the wave is:

   Options: 

    A. $y = 0.2 \sin [ 2\pi ( 6t + x/60 ) ]$  

    B. $y = 0.2 \sin [ \pi ( 6t + x/60 ) ]$  

    C. $y = 0.2 \sin [ 2\pi ( 6t - x/60 ) ]$  

    D. $y = 0.2 \sin [ \pi ( 6t - x/60 ) ]$

   Correct Answer: C

   Year: 2002

   Solution:  

    General equation for a plane progressive wave travelling along positive x-direction is given by $y = A \sin(\omega t - kx) = A \sin [ 2\pi (f t - x/\lambda) ]$.  

    Here, $v = 360$ m/s, $\lambda = 60$ m.  

    We know that, frequency $f = v / \lambda = 360 / 60 = 6$ Hz.  

    Substituting $A = 0.2$ m, $f = 6$ Hz, and $\lambda = 60$ m: $y = 0.2 \sin [ 2\pi (6t - x/60) ]$.

   Step Solution:

    1.  Identify Wave Direction: Traveling in the positive x-direction means the equation takes the form $(ft - x/\lambda)$ with a negative sign.

    2.  Calculate Frequency ($f$): Use the formula $f = v / \lambda = 360 / 60 = 6$ Hz.

    3.  Identify General Form: Use $y = A \sin [ 2\pi (ft - x/\lambda) ]$.

    4.  Substitute Values: Plug in $A = 0.2$, $f = 6$, and $\lambda = 60$.

    5.  Finalize Equation: $y = 0.2 \sin [ 2\pi (6t - x/60) ]$.

   Difficulty Level: Easy

   Concept Name: Standard Equation of Traveling Waves

   Shortcut Solution: Since the wave moves in the positive x-direction, the sign between $t$ and $x$ must be negative (eliminating A and B). Calculate frequency $f = 360/60 = 6$. Only option C has $6t$ and a negative sign.

Question 57

   Question: The equation of a wave is represented by $y = 10^{-4} \sin(100t - x/10)$ m, then the velocity of wave will be

   Options: 

    A. 100m/s  

    B. 4m/s  

    C. 1000m/s  

    D. 10m/s

   Correct Answer: C

   Year: 2001

   Solution:  

    Comparing the given equation with general equation, $y = a \sin(\omega t - kx)$, we get $\omega = 100$ and $k = 1/10$.  

    $v = \omega / k = 100 / (1/10) = 1000$ m/s.

   Step Solution:

    1.  Compare Equations: Match $y = 10^{-4} \sin(100t - 0.1x)$ with the standard form $y = A \sin(\omega t - kx)$.

    2.  Identify Angular Frequency ($\omega$): The coefficient of $t$ is $\omega = 100$ rad/s.

    3.  Identify Wave Number ($k$): The coefficient of $x$ is $k = 1/10$ rad/m.

    4.  Apply Velocity Formula: Wave velocity $v = \omega / k$.

    5.  Calculate: $v = 100 / (1/10) = 1000$ m/s.

   Difficulty Level: Easy

   Concept Name: Wave Velocity from Wave Equation

   Shortcut Solution: Wave velocity is simply the coefficient of $t$ divided by the coefficient of $x$. Calculation: $100 / (1/10) = 1000$.

Question 64

   Question: A transverse wave is represented by the equation $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$. For what value of $\lambda$, is the maximum particle velocity equal to two times the wave velocity?

   Options: 

    A. $\lambda = \pi y_0 / 2$  

    B. $\lambda = \pi y_0 / 3$  

    C. $\lambda = 2\pi y_0$  

    D. $\lambda = \pi y_0$

   Correct Answer: D

   Year: 1998

   Solution:  

    The given equation is $y = y_0 \sin \frac{2\pi}{\lambda} (vt - x)$.  

    Particle velocity $v_p = dy/dt = y_0 \cos [\frac{2\pi}{\lambda} (vt - x)] \cdot \frac{2\pi v}{\lambda}$.  

    $(v_p)_{max} = y_0 \cdot \frac{2\pi v}{\lambda}$.  

    According to the question: $y_0 \cdot \frac{2\pi v}{\lambda} = 2v \implies \lambda = \pi y_0$.

   Step Solution:

    1.  Identify Peak Particle Velocity: From the wave equation, the maximum particle velocity $(v_p)_{max}$ is $A\omega$.

    2.  Express $\omega$ in terms of given parameters: In this equation, $\omega = 2\pi v / \lambda$.

    3.  Set up the equation: $(v_p)_{max} = y_0 (2\pi v / \lambda)$.

    4.  Apply the condition: Set $y_0 (2\pi v / \lambda) = 2v$.

    5.  Solve for $\lambda$: $y_0 \cdot \pi / \lambda = 1 \implies \lambda = \pi y_0$.

   Difficulty Level: Medium

   Concept Name: Maximum Particle Velocity vs. Wave Velocity

   Shortcut Solution: Use the ratio formula: $(v_p)_{max} / v_{wave} = Ak$. Given ratio $= 2$. So, $y_0 \cdot k = 2$. Substituting $k = 2\pi / \lambda$ gives $y_0 (2\pi / \lambda) = 2$, which simplifies to $\lambda = \pi y_0$.

Question 70

   Question: The equation of a sound wave is $y = 0.0015 \sin (62.4x + 316t)$. The wavelength of this wave is

   Options: 

    A. 0.3 unit  

    B. 0.2 unit  

    C. 0.1 unit  

    D. cannot be calculated.

   Correct Answer: C

   Year: 1996

   Solution: Comparing it with the general equation of motion $y = A \sin 2\pi [ \frac{x}{\lambda} + \frac{t}{T} ]$, we get $\frac{2\pi}{\lambda} = 62.4$ or $\lambda = \frac{2\pi}{62.4} = 0.1 \text{ unit}$.

   Step Solution:

    1.  Identify the Wave Equation: Given $y = 0.0015 \sin(62.4x + 316t)$.

    2.  Extract the Wave Number ($k$): In the standard form $y = A \sin(kx + \omega t)$, the coefficient of $x$ is $k = 62.4$.

    3.  Use the Relation Formula: Wavelength $\lambda$ is related to $k$ by the formula $k = \frac{2\pi}{\lambda}$.

    4.  Rearrange for $\lambda$: $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14159}{62.4}$.

    5.  Final Calculation: $\lambda = \frac{6.283}{62.4} \approx \mathbf{0.1 \text{ unit}}$.

   Difficulty Level: Easy

   Concept Name: Wave Number and Wavelength Relationship

   Shortcut Solution: Wavelength $\lambda = \frac{2\pi}{\text{Coefficient of } x}$. Calculation: $6.28 / 62.4 \approx 0.1$.

Question 76

   Question: Which one of the following represents a wave?

   Options: 

    A. $y = A \sin(\omega t - kx)$  

    B. $y = A \cos(at - bx + c)$  

    C. $y = A \sin kx$  

    D. $y = A \sin \omega t$

   Correct Answer: B

   Year: 1994

   Solution: (a) represents a harmonic progressive wave in the standard form whereas (b) also represents a harmonic progressive wave, both travelling in the positive x-direction. In (b), $a$ is the angular velocity $\omega$ and $b$ is $k$; $c$ is the initial phase. (d) represents only S.H.M.

   Step Solution:

    1.  Define Wave Criteria: A traveling wave must be a function of the form $f(ax \pm bt)$.

    2.  Evaluate Option A: $y = A \sin(\omega t - kx)$ is a standard progressive wave.

    3.  Evaluate Option B: $y = A \cos(at - bx + c)$ is also a progressive wave (cosine form) with an initial phase $c$.

    4.  Evaluate Options C & D: Option C is a function of $x$ only (stationary/spatial) and Option D is a function of $t$ only (Simple Harmonic Motion).

    5.  Conclusion: While both A and B are waves, the source identifies B as the answer.

   Difficulty Level: Easy

   Concept Name: General Wave Equation Form

   Shortcut Solution: A traveling wave must contain both $x$ and $t$ variables within the trigonometric argument. Options A and B satisfy this; the source specifies B.

Question 84

   Question: The frequency of sinusodial wave $y = 0.40 \cos[2000t + 0.80]$ would be

   Options: 

    A. $1000\pi$ Hz  

    B. $2000$ Hz  

    C. $20$ Hz  

    D. $\frac{1000}{\pi}$ Hz

   Correct Answer: D

   Year: 1992

   Solution: Compare with the equation, $y = a \cos (2\pi \nu t + \phi)$. This give $2\pi \nu = 2000 \implies \nu = \frac{1000}{\pi}$ Hz.

   Step Solution:

    1.  Identify the standard form: Compare the given equation with $y = A \cos(\omega t + \phi)$.

    2.  Extract Angular Frequency ($\omega$): The coefficient of $t$ is $\omega = 2000$ rad/s.

    3.  Relate $\omega$ to Linear Frequency ($f$): Use the formula $\omega = 2\pi f$.

    4.  Set up the equation: $2\pi f = 2000$.

    5.  Calculate $f$: $f = \frac{2000}{2\pi} = \frac{1000}{\pi}$ Hz.

   Difficulty Level: Easy

   Concept Name: Frequency from Angular Frequency

   Shortcut Solution: The frequency $f$ is simply the coefficient of $t$ divided by $2\pi$. Calculation: $2000 / 2\pi = 1000/\pi$.

Question 92

   Question: Equation of progressive wave is given by $y = 4 \sin \left[ \pi \left( \frac{t}{5} - \frac{x}{9} \right) + \frac{\pi}{6} \right]$ where $y, x$ are in cm and $t$ is in seconds. Then which of the following is correct?

   Options: 

    A. $v = 5$ cm  

    B. $\lambda = 18$ cm  

    C. $a = 0.04$ cm  

    D. $f = 50$ Hz

   Correct Answer: B

   Year: 1988

   Solution: The standard equation of a progressive wave is $y = a \sin \left[ 2 \pi \left(\frac{t}{T} - \frac{x}{\lambda}\right) + \phi \right]$. The given equation can be written as $y = 4 \sin \left[ 2 \pi \left(\frac{t}{10} - \frac{x}{18}\right) + \frac{\pi}{6} \right]$. Therefore $a = 4$ cm, $T = 10$ s, $\lambda = 18$ cm and $\phi = \frac{\pi}{6}$.

   Step Solution:

    1.  Rewrite the equation: Multiply $\pi$ into the bracket to get $y = 4 \sin \left[ \frac{\pi t}{5} - \frac{\pi x}{9} + \frac{\pi}{6} \right]$.

    2.  Identify the Wave Number ($k$): Compare with $y = A \sin(\omega t - kx + \phi)$; here $k = \frac{\pi}{9}$.

    3.  Use Wavelength Formula: Use the relation $k = \frac{2\pi}{\lambda}$.

    4.  Substitute $k$: $\frac{2\pi}{\lambda} = \frac{\pi}{9}$.

    5.  Solve for $\lambda$: $\lambda = \frac{2\pi \times 9}{\pi} = 18$ cm.

   Difficulty Level: Easy

   Concept Name: Wave Parameter Extraction (Wavelength)

   Shortcut Solution: Wavelength $\lambda$ is $2\pi$ divided by the coefficient of $x$. Since the coefficient of $x$ is $\frac{\pi}{9}$, $\lambda = \frac{2\pi}{\pi/9} = 18$ cm.