Waves practice · Mixed difficulty

Option A ($A\omega$) is correct because it represents the maximum particle velocity in a wave, which is the product of the amplitude and angular frequency.

Option D ($f^{\prime}=f$) is correct because the changes in tension, density, and radius compensate each other, resulting in the same frequency.

Option A (50 Hz; 4 cm; 2 m/s) is correct because it accurately calculates the frequency, wavelength, and wave speed based on the given time and distance.

Option D ($\frac{2}{3}$) is correct because it is calculated using the formula for the ratio of transmitted wave amplitude to incident wave amplitude based on the speeds of sound in the two media.

Option A (5 cm) is correct because it is the calculated separation between successive nodes using the given parameters and the formula for wave speed on a string.

Option A is correct because the assertion and reason are both true, and the reason correctly explains why beats are not distinctly heard at those frequencies due to persistence of hearing.

Option C is correct because the assertion is true, but the phase difference between displacement and velocity in SHM is $\pi/2$, not $\pi$.

Option B ($50,150,250 \ldots \ldots$) is correct because closing one end of an open organ pipe halves the fundamental frequency and only odd harmonics are produced.

Option B (${27^0}C$) is correct because it is calculated using the relationship between the velocity of sound and temperature in an ideal gas.

Option D ($20\,\,mm$) is correct because it is calculated using the maximum audible frequency and the speed of sound.

Option A is correct because mechanical transverse waves require a medium with shearing properties, which gases lack.

Option D (All of these) is correct because the air column in a pipe closed at one end can resonate at the fundamental frequency and its odd harmonics.

Option C ($\frac{\pi}{2} + (\phi)$) is correct because to find the phase difference, both waves must be expressed as sine functions, and then the difference in their arguments is calculated.

Option B (178.2 Hz) is correct because it is the fundamental frequency calculated using the given density, elasticity, and strain of the steel wire.

Option A ($\frac{2(n+1)}{2 n+1}$) is correct because it represents the ratio of the frequencies of the nth overtone for an open and closed organ pipe of the same length.

Option B ($\frac{{2\pi }}{{3}}$) is correct because the resultant amplitude is equal to the individual amplitudes only when the phase difference is $\frac{{2\pi }}{{3}}$.

Option B (85, closed) is correct because the given frequencies are odd multiples of 85, indicating a closed pipe.

Option B ($64 \mathrm{~cm} \mathrm{~s}^{-1}$ in positive $x$-direction) is correct because the wave velocity is calculated from the coefficients of t and x in the wave equation.

Option A ($819^{\circ} \mathrm{C}$) is correct because the speed of sound is proportional to the square root of the temperature, and doubling the speed requires quadrupling the absolute temperature.

Option B ($V - v$,if $A = \lambda /2\pi$) is correct because the wave velocity equals the maximum particle velocity when the amplitude is equal to the wavelength divided by $2\pi$.

Option D (Both (1) and (2) options are correct) is correct because a Carnot engine operates between two temperatures with maximum efficiency and consists of two isothermal and two adiabatic processes.

Option C (It is the result of superposition of two waves of wavelength $3 {~m}$, frequency $60 {~Hz}$ each travelling with a speed of $180 {~m} / {s}$ in opposite direction) is correct because the given equation represents a stationary wave formed by the superposition of two waves with the specified wavelength, frequency, and speed.

Option C ($\frac{3}{8}\,\,m{s^{ - 1}}$) is correct because it is the velocity of the wave, calculated using the formula v = w/k.

The correct answer is Four times because the frequency of vibration of a string is directly proportional to the square root of the tension.

Option A (1) is correct because the tension in the string is calculated using the formula relating velocity, tension, and mass per unit length.

Option D (Quink's tube experiment is related with beats) is incorrect because Quink's tube experiment is related to interference, not beats.

Option C ($\frac{v}{9}$) is correct because it is derived from the Doppler effect equations for a moving observer and a stationary source.

The correct answer is A because Z₁ and Z₂ represent waves with the same frequency traveling in opposite directions, which is the condition for forming a stationary wave.

Option B (3) is correct because the beat frequency is the absolute difference between the frequencies of the two vibrations.

Option D (3 m, 60 Hz) is correct because the wavelength and frequency are derived from the given standing wave equation.