Waves — NEET Physics PYQ MCQs with Solutions

Solution: Wavelength of transverse pulse $\lambda = \frac{v}{f}$ As we know $v = \sqrt{\frac{T}{\mu}}$ (T = tension in the spring; μ = mass per unit length of the rope) From Eqs. (i) and (ii), we get $\lambda = \frac{1}{f}\sqrt{\frac{T}{\mu}}$ Therefore, $\lambda \propto \sqrt{T}$ So, for two different cases, we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{m_1 + m_2}{m_2}}$

Solution: Given, $a = 1$ m As $y = a\sin(kx - \omega t)$ $k = \frac{2\pi}{\lambda} = \frac{2\pi}{2\pi} = 1$ $\omega = 2\pi f = 2\pi \times 1 = 2\pi$ rad/s Therefore, $y = \sin(x - 2t)$ m

Solution: According to question, $y_1 = a\sin(\omega t + kx + 0.57)$ and $y_2 = a\cos(\omega t + kx) = a\sin\left(\frac{\pi}{2} + \omega t + kx\right)$ As phase difference, $\Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - 0.57 = 1.57 - 0.57 = 1$ rad

Solution: Velocity of a wave is given by $v = n\lambda$ where, $n$ = frequency of wave, $\lambda$ = wavelength of wave So, for two different cases: $v_1 = n_1\lambda_1$ $v_2 = n_2\lambda_2$ $\frac{\lambda_2}{\lambda_1} = \frac{\lambda_1 v_2}{v_1} = \lambda_1 \times \frac{3500}{350} = \lambda_1 \times 10$ [Since $n_1 = n_2$] Therefore, $\lambda_2 = 10\lambda_1$

Solution: Given, $y = A\sin(\omega t - kx)$ As we know that wave velocity is given by $v_w = \lambda/T = \omega\lambda/2\pi$ ...(i) and maximum particle velocity is given by $v_p = A\omega$ ...(ii) So, as Eq. (i) is equal to Eq. (ii), $A\omega = \frac{\omega\lambda}{2\pi}$, therefore $\lambda = 2\pi A$

Solution: Given, amplitude of wave, $A = 2$ cm direction = +ve x direction Velocity of wave $v = 128$ ms⁻¹ and length of string, $5\lambda = 4$ We know that, $k = \frac{2\pi}{\lambda} = \frac{2\pi \times 5}{4} = 7.85$ and $v = \frac{\omega}{k} = 128$ ms⁻¹ $\omega = v \times k = 128 \times 7.85 = 1005$ As, the wave travelling towards +x-axis is given by $y = A\sin(kx - \omega t)$ So, $y = 2\sin(7.85x - 1005t) = (0.02)$ m $\sin(7.85x - 1005t)$

Solution: $y = 0.25\sin(10\pi x - 2\pi t)$ Compare the above equation with $y = A\sin(kx - \omega t)$ As $\omega t$ and $kx$ have opposite sign, wave travels along positive x. As, $2\pi t = \omega t$ $\omega = 2\pi = 2\pi \nu$ $\nu = 1$ Hz Also, $kx = 10\pi x$ $k = \frac{2\pi}{\lambda} = 10\pi$ $\lambda = \frac{2\pi}{10\pi} = 0.2$ m

Solution: In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.

Solution: Sabine's formula for reverberation time is $T = 0.16\frac{V}{\sum as}$ where, V is volume of hall in m³. $\sum as$ = total absorption of the hall (room) So, $T \propto \frac{V}{s}$ For two different cases of reverberation: $\frac{T'}{T} = \frac{V'}{V} \times \frac{s}{s'} = \frac{(2)^3}{(2)^2} = \frac{8}{4} = 2$ Hence, $T' = 2T = 2 \times 1 = 2$ s

Solution: The given equation is $y(x,t) = 8.0\sin\left(0.5\pi x - 4\pi t - \frac{\pi}{4}\right)$ ...(i) Compare it with the standard wave equation $y = a\sin(kx - \omega t + \phi)$ ...(ii) where a is amplitude, k the propagation constant and ω the angular frequency. Comparing the Eqs. (i) and (ii), we have $k = 0.5\pi$, $\omega = 4\pi$ Speed of transverse wave, $v = \frac{\omega}{k} = \frac{4\pi}{0.5\pi} = 8$ m/s

Solution: Intensity of sound, $I = \frac{p}{4\pi r^2}$ or $I \propto \frac{1}{r^2}$. Therefore, $\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Solution: Converting $y_2$ to sine form: $y_2 = 10^{-6} \sin\left(100t + \frac{\pi}{50}x + \frac{\pi}{2}\right)$. Phase difference = $\frac{\pi}{2} - 0.5 = 1.57 - 0.5 = 1.07$ rad

Solution: For positive x-direction wave: $y = a \sin(\omega t - kx)$. With $\omega = 2\pi\frac{v}{\lambda} = 2\pi \cdot 6$ and $k = \frac{2\pi}{\lambda} = \frac{2\pi}{60}$

Solution: Comparing with standard form $y = a \sin(\omega t - kx)$: $\omega = 100$, $k = \frac{1}{10}$. Velocity $v = \frac{\omega}{k} = \frac{100}{1/10} = 1000$ m/s

Solution: When a wave enters from one medium to another, its frequency remains unchanged because it is determined by the source. Wavelength, intensity and velocity all change in the new medium.

Solution: Frequency of vibrations of string is $n = \frac{1}{2l}\sqrt{\frac{g}{\mu}}$. This does not depend on mass. From $n_A = 2n_B$: $\frac{1}{l_A} = \frac{2}{l_B}$, so $l_B = 4l_A$

Solution: Time from max to zero displacement = $\frac{T}{4}$. So $T = 0.170 \times 4 = 0.680$ s. Frequency $f = \frac{1}{T} = \frac{1}{0.680} = 1.47$ Hz

Solution: Maximum particle velocity: $u_{max} = y_0 \omega = y_0 \cdot \frac{2\pi}{T}$. Wave velocity: $v_w = \frac{\lambda}{T}$. Setting $u_{max} = 2v_w$: $y_0 \cdot \frac{2\pi}{T} = 2 \cdot \frac{\lambda}{T}$, giving $\lambda = \pi y_0$

Solution: Comparing with $y = a \sin(\omega t + kx)$: $k = 62.4$. Since $k = \frac{2\pi}{\lambda}$, we have $\lambda = \frac{2\pi}{62.4} = \frac{2\pi}{62.4} = 0.1$ unit

Solution: The presence of water vapours reduces the density of air, which increases the velocity of sound: $v_m = \sqrt{\frac{\rho_d}{\rho_m}} v_d > v_d$ since $\rho_m < \rho_d$

Solution: Frequency = $\frac{3600}{120} = 30$ Hz. Wavelength = $\frac{v}{f} = \frac{760}{30} = 25.3$ m

Solution: Wavelength = $\frac{v}{f} = \frac{1.7 \times 10^3}{4.2 \times 10^6} = 4 \times 10^{-4}$ m

Solution: Two waves are coherent when they have the same frequency, amplitude and constant phase difference.

Solution: Comparing with $y = a \sin 2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)$: $v = 64$ and $\lambda = 3.2$. Frequency = $\frac{v}{\lambda} = \frac{64}{3.2} = 20$ Hz

Solution: Velocity of sound: $c_s = \sqrt{\frac{\gamma p}{\rho}}$. RMS velocity: $c = \sqrt{\frac{3p}{\rho}}$. Therefore, $c_s = \frac{c}{\sqrt{3}}$

Solution: A wave equation must have both space (x) and time (t) variables with opposite signs. The form $y = A \sin(\omega t - kx)$ represents a wave travelling in the positive x-direction.

Solution: Comparing with $y = a \cos(2\pi f \cdot t + \phi)$: $2\pi f = 2000$, so $f = \frac{2000}{2\pi} = \frac{1000}{\pi}$ Hz. But if interpreted as $2\pi f = 2000$, then checking: $2\pi \times 1000 = 2000\pi \neq 2000$. Actually $f = \frac{2000}{2\pi} \approx 318$ Hz. However, standard answer assumes $\omega = 2000\pi$ rad/s, giving $f = 1000/\pi$ Hz or if 2000 is already in rad/s form, we interpret differently. The answer given is 1000 Hz implying interpretation as $\omega = 2000\pi$ rad/s.

Solution: In longitudinal waves, energy is propagated along with the wave motion without any net transport of the mass of the medium.

Solution: Comparing with $y = a \sin 2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)$: $\omega = 15$, $k = 3$. Speed = $\frac{\omega}{k} = \frac{15}{3} = 5$ m/s

Solution: Phase change for path difference: $\Delta\phi = \frac{2\pi}{\lambda} \cdot \Delta x$. So $1.6\pi = \frac{2\pi}{\lambda} \times 0.4$, giving $\lambda = 0.5$ m. Frequency = $\frac{v}{\lambda} = \frac{330}{0.5} = 660$ Hz