JEE Physics Waves MCQs Set 18

Answer: (b) all four statements (i), (ii), (iii) and (iv)

Explanation:
1. \( y_{max}(5)=4\sin\frac{\pi}{3}=2\sqrt3\ \text{cm} \) (i is true).
2. Nodes: \( \frac{\pi x}{15}=n\pi\Rightarrow x=15n \), and \( 15n\le600 \) gives n = 0 to 40 (ii is true).
3. Velocity \( \propto\sin(96\pi t) \) and \( 96\pi\times0.25=24\pi \), so the velocity is zero (iii is true).
4. \( 2\sin2\pi\left(\frac{x}{30}+48t\right)+2\sin2\pi\left(\frac{x}{30}-48t\right)=4\sin\frac{\pi x}{15}\cos96\pi t \) (iv is true).

Teacher's Note:
1. Nodes are where sin(πx/15) = 0.
2. Velocity of a particle in a standing wave is zero when cos(ωt) is at an extremum.

Answer: (d) (i), (ii) and (iii) only

Explanation:
1. The reflected wave travels in the opposite direction and is inverted at the fixed end: \( \xi_r=-10\sin(80\pi t+4\pi x) \) (i true).
2. Standing wave amplitude \( 2\times10 \), twice the original (ii true).
3. \( \lambda=\frac{2\pi}{4\pi}=0.5\ \text{m} \) (iii true).
4. Length 1 m = 2λ gives 4 loops and 5 nodes including the ends, not 3 (iv false).

Teacher's Note:
1. Reflection from a fixed end inverts the wave.
2. Number of nodes including the ends is one more than the number of loops.

Answer: (c) 2.5 cm²

Explanation:
1. The pulse is a triangle with vertices at x = -4, x = 0 (height 1 cm) and x = 1 cm.
2. Base = 5 cm and height = 1 cm, so \( \text{Area}=\tfrac12\times5\times1=2.5\ \text{cm}^2 \).

Teacher's Note:
1. The pulse rises from -4 to 0 and falls from 0 to 1.
2. Area of a triangle is half base times height.

Answer: (d) 0.25 cm

Explanation:
1. \( v=\sqrt{T/\mu}=\sqrt{5/0.05}=10\ \text{m/s}=1000\ \text{cm/s} \).
2. In 0.01 s the pulse moves 10 cm, so \( y(7,0.01)=y_0(7-10)=y_0(-3) \).
3. For \( -4\lt x\le0 \): \( y_0=\frac{-3}{4}+1=0.25\ \text{cm} \)

Teacher's Note:
1. A pulse moving right satisfies y(x,t) = y0(x − vt).
2. Use the segment of the initial shape that contains the shifted position.

Answer: (b) -250 cm/s

Explanation:
1. Pulse shift in 0.015 s: \( 1000\times0.015=15\ \text{cm} \), so the point corresponds to \( x-vt=13-15=-2 \) on the initial shape.
2. There the slope is \( \frac{dy}{dx}=\frac14 \).
3. \( v_y=-v\frac{\partial y}{\partial x}=-1000\times\frac14=-250\ \text{cm/s} \)

Teacher's Note:
1. For y = f(x − vt), the particle velocity is −v times the slope.
2. The slope on the left segment is +1/4.

Answer: (d) \( 10.8\pi \)

Explanation:
1. \( v=\sqrt{\frac{900}{0.09}}=100\ \text{m/s},\quad \omega=2\pi\times120=240\pi,\quad k=\frac{\omega}{v}=2.4\pi\ \text{m}^{-1} \).
2. The transverse component of tension is \( T\sin\theta\approx T\left|\frac{\partial y}{\partial x}\right| \), with maximum slope \( Ak=0.005\times2.4\pi \).
3. \( F_{y,max}=900\times0.005\times2.4\pi=10.8\pi\ \text{N} \)

Teacher's Note:
1. For small slopes sin θ ≈ tan θ = dy/dx.
2. The maximum slope of a sine wave is Ak.

Answer: (d) \( \frac12 \)

Explanation:
1. Power \( =-T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t} \) is zero when the slope is zero, i.e. at a crest or trough.
2. There the displacement has its maximum magnitude, equal to the amplitude \( A=\tfrac12\ \text{cm} \).

Teacher's Note:
1. Power transfer needs both slope and particle velocity to be non-zero.
2. At a crest the slope is zero and the particle is momentarily at rest.

Answer: (b) \( \Delta\phi=\pi \) between the reflected and transmitted waves

Explanation:
1. Linear densities: \( \mu_1=\rho\pi(2r)^2=4\pi\rho r^2 \), \( \mu_2=9\rho\pi r^2 \); \( \mu_2>\mu_1 \): BC is the denser string, with lower wave speed.
2. Reflection at a denser medium gives a phase change of \( \pi \) between incident and reflected waves; the transmitted wave is in phase with the incident wave.
3. Hence reflected and transmitted waves differ by \( \pi \).

Teacher's Note:
1. Going from a light to a heavy string, the reflected wave is inverted.
2. The transmitted wave always keeps the phase of the incident wave.

Answer: (d) 3 : 2

Explanation:
1. \( v_1=\sqrt{T/\mu_1},\ v_2=\sqrt{T/\mu_2}\Rightarrow\frac{v_1}{v_2}=\sqrt{\frac{9}{4}}=\frac32 \).
2. The reflected wave stays in string AB with the same speed as the incident wave, so \( \lambda_r=\lambda_i=\frac{v_1}{f} \) and \( \lambda_t=\frac{v_2}{f} \).
3. \( \lambda_r:\lambda_t=v_1:v_2=3:2 \)

Teacher's Note:
1. Frequency is common to all three waves.
2. Wavelength is proportional to the speed in the string.

Answer: (a) 1 : 4

Explanation:
1. \( A_r=\frac{v_2-v_1}{v_2+v_1}A_i=\frac{2-3}{2+3}A_i=-\frac15A_i \) (taking \( v_1:v_2=3:2 \)).
2. \( A_t=\frac{2v_2}{v_1+v_2}A_i=\frac{4}{5}A_i \).
3. So \( |A_r|:A_t=1:4 \).

Teacher's Note:
1. Use the reflection and transmission coefficients in terms of wave speeds.
2. The negative sign of A_r indicates the phase change of π.

Answer: (d) 5

Explanation:
1. \( \mu=\frac{1.0\times10^{-3}}{0.20}=5\times10^{-3}\ \text{kg/m} \).
2. \( v=\sqrt{\frac{0.5}{5\times10^{-3}}}=10\ \text{m/s},\quad \lambda=\frac{v}{f}=0.1\ \text{m} \).
3. Successive nodes are \( \frac{\lambda}{2}=5\ \text{cm} \) apart.

Teacher's Note:
1. Distance between adjacent nodes is half a wavelength.
2. λ = v/f.

Answer: (b) (ii) and (iii) only

Explanation:
1. \( k=62.8\Rightarrow\lambda=\frac{2\pi}{62.8}=0.1\ \text{m} \). The fifth harmonic has \( L=\frac{5\lambda}{2}=0.25\ \text{m} \) (ii true).
2. It has 5 loops and 6 nodes counting the ends, so (i) is false.
3. Midpoint x = 0.125 m: \( \sin(62.8\times0.125)=\sin(2.5\pi)=1 \), so the amplitude is 0.01 m (iii true).
4. \( f_5=\frac{628}{2\pi}=100\ \text{Hz} \), so the fundamental is 20 Hz (iv false).

Teacher's Note:
1. Five loops means six nodes including both ends.
2. The fundamental is one fifth of the fifth harmonic.

Answer: (c) (i), (iii) and (iv) only

Explanation:
1. All four satisfy \( \omega/k=100\ \text{m/s} \).
2. The end at x = 3 m vibrates, so it must not be a node: \( \sin(kx)\neq0 \) at x = 3.
3. \( \text{(i)}\ \sin\frac{\pi}{2}=1,\ \text{(ii)}\ \sin\pi=0,\ \text{(iii)}\ \sin\frac{5\pi}{2}=1,\ \text{(iv)}\ \sin\frac{15\pi}{2}=-1 \).
4. So (ii) is excluded; (i), (iii) and (iv) are possible.

Teacher's Note:
1. The driven end must have non-zero amplitude.
2. Check that ω/k equals the wave speed for every candidate.

Answer: (d) (ii) and (iv) only

Explanation:
1. The speed at a point is \( v=\sqrt{T(y)/\mu} \), fixed by the tension at that point, so the speed does not depend on frequency or wavelength (ii true).
2. At the midpoint the two pulses have equal speeds but move in opposite directions (one down, one up), so their velocities (vectors) are not the same: (i) is false.
3. Tension decreases going down, so Pulse 1 slows down; frequency is constant, so its wavelength becomes shorter at A (iii false).
4. Both pulses cover the same speed profile between O and A, so \( T_{AO}=T_{OA} \) (iv true).

Teacher's Note:
1. Velocity is a vector: direction matters as well as speed.
2. Wavelength = speed ÷ frequency, and the frequency is constant.

Answer: (b) a standing wave with nodes at \( x=\left(n+\tfrac12\right)\dfrac{\lambda}{2},\ n=0,1,2,\dots \)

Explanation:
1. \( y=2A\sin\omega t\cos kx \), a standing wave.
2. Nodes at \( \cos kx=0\Rightarrow kx=\left(n+\tfrac12\right)\pi \), i.e. \( x=\left(n+\tfrac12\right)\frac{\lambda}{2} \).

Teacher's Note:
1. Use sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2).
2. Nodes are where the spatial factor is zero.

Answer: (c) \( 2\sqrt2 \) s

Explanation:
1. At height y above the free end, \( v=\sqrt{gy} \).
2. \( t=\int_0^L\frac{dy}{\sqrt{gy}}=2\sqrt{\frac{L}{g}}=2\sqrt{\frac{20}{10}}=2\sqrt2\ \text{s} \)

Teacher's Note:
1. Tension at height y is μgy because it supports the weight below.
2. Integrate dt = dy/v.

Answer: (d) 157.5 m/s

Explanation:
1. The fundamental is \( f_1=420-315=105\ \text{Hz} \).
2. \( v=2Lf_1=2\times0.75\times105=157.5\ \text{m/s} \)

Teacher's Note:
1. Consecutive harmonics differ by the fundamental frequency.
2. v = 2Lf1 for a string fixed at both ends.

Answer: (c) 162 Hz and 3 nodes

Explanation:
1. Tension \( T=10\times9.8=98\ \text{N} \).
2. \( v_{Al}=\sqrt{\frac{98}{2.6\times10^{-3}}}=194.2\ \text{m/s} \), \( v_{St}=\sqrt{\frac{98}{1.04\times10^{-2}}}=97.1\ \text{m/s} \).
3. Fundamental of each segment: aluminium \( \frac{194.2}{2(0.6)}=161.8\ \text{Hz} \), steel (0.9 m) \( \frac{97.1}{2(0.9)}=53.9\ \text{Hz} \).
4. The lowest common frequency is \( 161.8=3\times53.9 \): 1 loop in aluminium and 3 in steel, so \( f\approx162\ \text{Hz} \).
5. Nodes excluding the ends: 0 in aluminium, 2 in steel, plus the joint = 3.

Teacher's Note:
1. The steel segment is 0.9 m because 1.5 − 0.6 = 0.9.
2. A segment with n loops has n − 1 nodes inside it.

Answer: (a) a decrease of 5 Hz

Explanation:
1. \( f\propto\sqrt T:\ \sqrt{1.44}=1.2\Rightarrow0.2f=6\Rightarrow f=30\ \text{Hz} \).
2. Length +20% at constant tension: \( f'=\frac{f}{1.2}=25\ \text{Hz} \).
3. The frequency decreases by 5 Hz.

Teacher's Note:
1. Frequency is inversely proportional to length.
2. First find the original frequency from the tension change.

Answer: (a) \( 2\pi/5 \) N

Explanation:
1. Fundamental: \( y=a\sin\frac{\pi x}{L}\cos\omega t \) with \( a=0.02\ \text{m},\ L=0.5\ \text{m} \).
2. Maximum slope at the end: \( \frac{a\pi}{L}=\frac{0.02\pi}{0.5}=0.04\pi \).
3. \( F_y=T\times\text{slope}=10\times0.04\pi=0.4\pi=\frac{2\pi}{5}\ \text{N} \)

Teacher's Note:
1. For small slopes the vertical component of tension is T × slope.
2. The slope is greatest at the fixed end.

Answer: (a) 300 Hz

Explanation:
1. Mode n has displacement amplitude \( \left|\sin\frac{n\pi x}{L}\right| \); at \( x=L/4 \) this is \( |\sin(n\pi/4)| \).
2. It equals 1 for \( n=2,6,10,\dots \) (n = 1 and 3 give 0.707, n = 4 gives 0, n = 5 gives 0.707). The lowest mode is n = 2 with 100 Hz, so \( f_1=50\ \text{Hz} \).
3. The next such mode is n = 6: \( f=6\times50=300\ \text{Hz} \).

Teacher's Note:
1. The point at L/4 is an antinode only for n = 2, 6, 10, ...
2. Fundamental = 100/2 = 50 Hz.

Answer: (a) 225 N

Explanation:
1. \( \mu=\frac{4\times10^{-3}}{0.90}=4.44\times10^{-3}\ \text{kg/m} \).
2. \( v=2Lf=2\times0.9\times125=225\ \text{m/s} \).
3. \( T=\mu v^2=4.44\times10^{-3}\times225^2=225\ \text{N} \)

Teacher's Note:
1. Fundamental of a fixed-fixed string: f = v/2L.
2. T = μv².

Answer: (c) 0.66 m, 0.33 m, 0.22 m

Explanation:
1. Fundamental frequency is inversely proportional to length, so \( \ell_1:\ell_2:\ell_3=1:\tfrac12:\tfrac13=6:3:2 \).
2. Total = 11 parts = 1.21 m, so one part = 0.11 m.
3. Lengths: 0.66 m, 0.33 m, 0.22 m.

Teacher's Note:
1. Lengths are in the inverse ratio of the frequencies.
2. Add the ratio numbers to find the size of one part.

Answer: (b) \( \tfrac14m\omega^2a_{max}^2 \)

Explanation:
1. Take \( y=a\sin\frac{\pi x}{L}\cos\omega t \); the velocity is \( -a\omega\sin\frac{\pi x}{L}\sin\omega t \).
2. Maximum KE: \( \int_0^L\tfrac12\frac mL a^2\omega^2\sin^2\frac{\pi x}{L}dx=\tfrac12\frac mLa^2\omega^2\cdot\frac L2=\tfrac14m\omega^2a^2 \).

Teacher's Note:
1. The average of sin² over the length is 1/2.
2. The maximum KE occurs when the string passes through its straight position.

Answer: (d) \( \tfrac18m\omega^2a_{max}^2 \)

Explanation:
1. The kinetic energy varies as \( \sin^2\omega t \) with maximum \( \tfrac14m\omega^2a^2 \).
2. The time average of \( \sin^2\omega t \) is 1/2, so \( \langle K\rangle=\tfrac18m\omega^2a^2 \).

Teacher's Note:
1. The average of sin² over a full cycle is 1/2.
2. Half of the total energy is kinetic on average.