JEE Physics Waves MCQs Set 06

Answer: (d) 600 Hz

Explanation:
1. The frequency of a wave is fixed by the source and does not change when the wave passes from one medium to another.
2. Only the speed and the wavelength change (\( \lambda = v/f \)).
3. So the observer in air records 600 Hz.

Teacher's Note:
1. Frequency belongs to the source, speed belongs to the medium.
2. When a wave crosses a boundary, v and \( \lambda \) change together so that \( f = v/\lambda \) stays the same.

Answer: (a) 1.05 cm

Explanation:
1. Second resonance minus first resonance gives half a wavelength: \( \lambda/2 = 70.2 - 22.7 = 47.5 \) cm, so \( \lambda = 95 \) cm.
2. First resonance: \( L_1 + e = \lambda/4 = 23.75 \) cm.
3. End correction \( e = 23.75 - 22.7 = 1.05 \) cm.

Teacher's Note:
1. The difference of two successive resonance lengths removes the end correction.
2. Then use \( L_1 + e = \lambda/4 \) to find e.

Answer: (b) sound waves require a material medium for propagation

Explanation:
1. Sound is a mechanical wave and needs a material medium such as air.
2. Space between the moon and the earth is almost a vacuum, so the sound cannot travel across it.

Teacher's Note:
1. Light and radio waves can cross a vacuum but sound cannot.
2. Mechanical waves always need a medium.

Answer: (c) Resonance

Explanation:
1. In a sonometer the wire is made to vibrate by a tuning fork.
2. When the natural frequency of the wire equals that of the fork, the wire vibrates with maximum amplitude, which is resonance.

Teacher's Note:
1. The paper rider falls off at the loop where resonance happens.
2. Sonometer is used to find frequency of a fork or to verify the laws of vibrating strings.

Answer: (c) \( 330\sqrt{2} \) m/s

Explanation:
1. For an ideal gas \( v = \sqrt{\gamma RT/M} \), so v depends only on the absolute temperature, not on pressure.
2. Doubling T multiplies v by \( \sqrt{2} \).
3. Halving the pressure changes density and pressure together, so it has no effect. \( v' = 330\sqrt{2} \) m/s.

Teacher's Note:
1. The ratio P/ρ is what matters, and it depends only on T.
2. So pressure alone never changes the speed of sound in a gas.

Answer: (c) \( E_2 = 4E_1 \)

Explanation:
1. Both waves have the same amplitude A, but the second has twice the angular frequency.
2. Energy of a vibrating string is proportional to \( A^2\omega^2 \) (the mass is the same).
3. \( E_2/E_1 = (2\omega)^2/\omega^2 = 4 \).

Teacher's Note:
1. Energy in vibration depends on the square of amplitude and the square of frequency.
2. The mode number only matters through the frequency.

Answer: (d) 25 : 4

Explanation:
1. Energy density is proportional to \( A^2 \) when frequency and medium are the same.
2. Ratio \( = 5^2 : 2^2 = 25 : 4 \).

Teacher's Note:
1. Energy and intensity go as the square of amplitude.
2. Squaring the ratio 5 : 2 gives 25 : 4.

Answer: (a) 1.07 radians

Explanation:
1. Write the cosine as a sine: \( y_2 = 10^{-6}\sin\left[100t + \dfrac{x}{50} + \dfrac{\pi}{2}\right] \).
2. Phase of \( y_1 \) is \( \theta + 0.5 \) and phase of \( y_2 \) is \( \theta + \pi/2 \).
3. Difference \( = \pi/2 - 0.5 = 1.5708 - 0.5 = 1.07 \) rad.

Teacher's Note:
1. Convert both waves to the same trigonometric function first.
2. Then subtract the constant phase angles.

Answer: (b) 1450 m/s

Explanation:
1. \( v = \sqrt{B/\rho} \).
2. \( v = \sqrt{2.1\times10^9/1000} = \sqrt{2.1\times10^6} \approx 1449 \) m/s.
3. This is nearest to 1450 m/s.

Teacher's Note:
1. For liquids the speed of sound is \( \sqrt{B/\rho} \).
2. Keep the bulk modulus in pascal: 2100 MPa = \( 2.1\times10^9 \) Pa.

Answer: (c) 12.0 cm

Explanation:
1. Wavelength \( \lambda = v/f = 350/500 = 0.7 \) m.
2. Path difference \( \Delta x = \dfrac{\lambda}{2\pi}\times\Delta\phi = \dfrac{0.7}{360^\circ}\times 60^\circ = 0.1167 \) m.
3. This is about 12 cm.

Teacher's Note:
1. A phase difference of \( 360^\circ \) corresponds to one wavelength.
2. So \( 60^\circ \) corresponds to \( \lambda/6 \).

Answer: (a) 102 Hz

Explanation:
1. The unknown fork is either 102 Hz or 98 Hz.
2. Loading a fork with wax lowers its frequency.
3. If it were 98 Hz, loading would make it lower still and the beats would increase to 3 or more. If it is 102 Hz, loading lowers it to 101 Hz and the beats fall to 1.
4. So the frequency is 102 Hz.

Teacher's Note:
1. Loading lowers frequency, filing raises it.
2. Check which starting value gives fewer beats after loading.

Answer: (c) 1 s^-1

Explanation:
1. Compare with \( y = A\sin(kx - \omega t) \): \( \omega = 2\pi \) rad/s.
2. \( f = \omega/2\pi = 1 \) s^-1.

Teacher's Note:
1. Expand the bracket first so that the coefficient of t is \( \omega \).
2. Then \( f = \omega/2\pi \).

Answer: (c) 5400 km

Explanation:
1. Let the distance be d. Time for S wave \( = d/6 \), time for P wave \( = d/9 \) (d in km, time in s).
2. \( d/6 - d/9 = 300 \Rightarrow d/18 = 300 \).
3. \( d = 5400 \) km.

Teacher's Note:
1. Both waves travel the same distance, so the faster one arrives first.
2. The time gap equals \( d(1/v_S - 1/v_P) \).

Answer: (a) more frequency, less wavelength

Explanation:
1. Relative approach raises the observed frequency: \( f' = f\,\dfrac{v+v_L}{v-v_S} \) is greater than f.
2. The source moving towards the listener crowds the waves in the medium, so the wavelength \( \lambda' = (v - v_S)/f \) is shorter than \( v/f \).

Teacher's Note:
1. Approach means higher pitch.
2. The motion of the source changes the wavelength in the medium, while the motion of the observer changes only the rate at which crests are met.

Answer: (d) 64 cm s^-1 in +x direction

Explanation:
1. Expand: \( y = 4\sin(4\pi t - \pi x/16) \), so \( \omega = 4\pi \) and \( k = \pi/16 \).
2. \( v = \omega/k = 4\pi\times16/\pi = 64 \) cm/s.
3. The terms \( \omega t \) and \( kx \) have opposite signs, so the wave travels in the +x direction.

Teacher's Note:
1. Opposite signs of t and x mean travel along +x.
2. Speed is the ratio of the coefficients of t and x.

Answer: (a) maximum

Explanation:
1. A phase difference of \( 2\pi \) is equivalent to zero phase difference, so the waves are in phase.
2. The resultant amplitude \( \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos\phi} \) with \( \cos 2\pi = 1 \) gives \( a_1 + a_2 \), the maximum.

Teacher's Note:
1. Phase differences that are even multiples of \( \pi \) give constructive interference.
2. Odd multiples of \( \pi \) give destructive interference.

Answer: (c) \( \dfrac{2}{5}\cos^{-1}\left(\dfrac{3}{5}\right) \) s

Explanation:
1. The received frequency is maximum or minimum when the line from the source to the observer is tangent to the circle, because only then the whole speed of the source is along the line of sight.
2. For such a tangent point, the radius to the source makes an angle \( \theta \) with the line to the observer where \( \cos\theta = 3/5 \).
3. The two tangent points are separated by an angle \( 2\theta \) on the side facing the observer. The source takes \( t = 2\theta/\omega = \dfrac{2}{5}\cos^{-1}(3/5) \) s to go from the one to the other.

Teacher's Note:
1. Frequency is extreme when the source velocity points along the line to the observer.
2. The tangent from the observer to the circle gives \( \cos\theta = r/d = 3/5 \).

Answer: (d) \( \dfrac{A}{2},\ \dfrac{\pi}{k} \)

Explanation:
1. \( \sin^2\theta = \dfrac{1 - \cos2\theta}{2} \), so \( y = \dfrac{A}{2} - \dfrac{A}{2}\cos[2(kx - \omega t + \phi)] \).
2. The oscillating part has amplitude \( A/2 \) and wave number \( 2k \).
3. Wavelength \( = 2\pi/(2k) = \pi/k \).

Teacher's Note:
1. Use the double angle identity to turn a squared sine into a plain cosine.
2. The wave number doubles, so the wavelength halves.

Answer: (d) 15 m s^-1

Explanation:
1. Apparent frequency \( f' = f\,\dfrac{v}{v - V} \).
2. For the limiting case \( 10000 = 9500\times\dfrac{300}{300 - V} \Rightarrow 300 - V = 285 \).
3. \( V = 15 \) m/s.

Teacher's Note:
1. The source moving towards the observer increases the frequency.
2. The listener hears the whistle only while \( f' \le 10{,}000 \) Hz.

Answer: (b) 0.4 %

Explanation:
1. For a receding source, \( \dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c} \).
2. \( \dfrac{1.2\times10^6}{3\times10^8} = 0.004 = 0.4\% \).

Teacher's Note:
1. This is the Doppler red shift.
2. For speeds much smaller than c the fractional change in wavelength is v/c.

Answer: (d) 16

Explanation:
1. Increase in level \( = 10\log_{10}(P_2/P_1) \) dB.
2. \( = 10\log_{10}(400/10) = 10\log_{10}40 = 10\times1.602 \approx 16 \) dB.

Teacher's Note:
1. Decibel difference depends only on the ratio of powers.
2. Remember \( \log_{10}4 = 0.6 \) and \( \log_{10}10 = 1 \).

Answer: (a) 1

Explanation:
1. \( \nu = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}} \) and \( \mu = \rho\pi r^2 \), so \( \nu \propto \dfrac{1}{Lr} \) for the same T and \( \rho \).
2. \( \nu_1/\nu_2 = \dfrac{1/(L\cdot2r)}{1/(2L\cdot r)} = 1 \).

Teacher's Note:
1. Mass per unit length depends on the square of the radius.
2. Doubling the radius and halving the length leaves the product Lr unchanged.

Answer: (a) 200 s^-1

Explanation:
1. Let the open pipe fundamental be f. Its first overtone is 2f.
2. The closed pipe of the same length has fundamental f/2, and its second overtone is the fifth harmonic, \( 5\times f/2 = 2.5f \).
3. \( 2.5f - 2f = 100 \Rightarrow f = 200 \) Hz.

Teacher's Note:
1. A closed pipe has only odd harmonics.
2. Its second overtone is the 5th harmonic.

Answer: (d) 80 m

Explanation:
1. \( k = 0.157 = \pi/20 \), so \( \lambda = 2\pi/k = 40 \) m.
2. For the 4th harmonic \( L = 4\lambda/2 = 2\lambda \).
3. \( L = 80 \) m.

Teacher's Note:
1. In the nth harmonic \( L = n\lambda/2 \).
2. Here \( 0.157 \) is \( \pi/20 \), which gives the wavelength quickly.

Answer: (a) 9 : 4

Explanation:
1. For a point source \( I \propto 1/r^2 \).
2. \( I_P/I_Q = (3/2)^2 = 9/4 \).

Teacher's Note:
1. Intensity falls with the square of the distance.
2. The nearer point has the larger intensity.