Waves and Wave Properties — 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: 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: 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: 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: Frequency = $\frac{3600}{120} = 30$ Hz. Wavelength = $\frac{v}{f} = \frac{760}{30} = 25.3$ 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: 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

Solution: Velocity of wave: $v = \sqrt{\frac{T}{\mu}}$ where $\mu = \frac{0.035}{5.5}$ kg/m. $v = \sqrt{\frac{77}{0.035/5.5}} = \sqrt{\frac{77 \times 5.5}{0.035}} = 110$ m/s

Solution: Rewriting: $y = 4 \sin 2\pi\left(\frac{t}{10} - \frac{x}{18} + \frac{1}{6}\right)$. Comparing with standard form: $T = 10$ s, $\lambda = 18$ cm

Solution: For each part: $n_1 = \frac{v}{2l_1} \Rightarrow l_1 = \frac{v}{2n_1}$, $n_2 = \frac{v}{2l_2} \Rightarrow l_2 = \frac{v}{2n_2}$, $n_3 = \frac{v}{2l_3} \Rightarrow l_3 = \frac{v}{2n_3}$. For complete wire: $n = \frac{v}{2l} \Rightarrow l = \frac{v}{2n}$. Since $l = l_1 + l_2 + l_3$, we have $\frac{v}{2n} = \frac{v}{2n_1} + \frac{v}{2n_2} + \frac{v}{2n_3}$, which simplifies to $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

Solution: Let unknown frequency be $f$. Given: $|f - 250| = 4$ and $|2f - 513| = 5$. If $f = 254$: beats = $|254 - 250| = 4$ ✓ and $|2 \times 254 - 513| = |508 - 513| = 5$ ✓. Therefore, unknown frequency is 254 Hz.

Solution: Fundamental frequency: $\nu = \frac{1}{2l}\sqrt{\frac{T}{m}}$, so $\nu l = k$ (constant). For segments: $l_1 = \frac{k}{\nu_1}$, $l_2 = \frac{k}{\nu_2}$, $l_3 = \frac{k}{\nu_3}$. Original length: $l = l_1 + l_2 + l_3 = \frac{k}{\nu_1} + \frac{k}{\nu_2} + \frac{k}{\nu_3}$. Since $l = \frac{k}{\nu}$: $\frac{1}{\nu} = \frac{1}{\nu_1} + \frac{1}{\nu_2} + \frac{1}{\nu_3}$

Solution: Comparing with $y = a\sin(2\pi ft)$: $f_1 = 300$ Hz, $f_2 = 304$ Hz. Beat frequency = $|f_2 - f_1| = 4$ beats/s. Amplitude ratio $a_1:a_2 = 4:5$. Intensity ratio $I_{max}:I_{min} = \frac{(a_1+a_2)^2}{(a_1-a_2)^2} = \frac{(9)^2}{(1)^2} = 81:1$.