CBSE Class 11 Physics Wave Motion Advanced Problems

1. In a stationary wave that is formed as a result of reflection of waves from an obstacle, the ratio of the amplitude at an antinode to the amplitude at node is 6. What percentage of energy is transmitted.

2. A standing wave y = a sin kx cos ωt is maintained in a homogeneous rod with cross-sectional area S and density ρ. Find the total mechanical energy confined between the sections corresponding to the adjacent nodes.

3. A siren creates a sound level of 60 dB at a location 500 m from the speaker. The siren is powered by a battery that delivers a total energy of 1.0 kJ. Assuming that the efficiency of siren is 30%, determine the total time for which the siren can sound.

4. A stationary observer receives a sound from a source of frequency 2000 Hz moving with a constant velocity. The apparent frequency varies with time as shown in figure. Find

(a) speed of source (vs)
(b) maximum value of apparent frequency fm. (speed of 1800 sound is v = 300 m/s)

5. A source of sonic oscillations with frequency f Hz and a receiver are located at the same point. At t = 0, the source starts receding from the receiver with constant acceleration a. Assuming the velocity of sound to be c, find the oscillation frequency registered by stationary receiver t0 second after the start up of the motion.

6. A metallic rod of length 1 m is rigidly clamped at its mid point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid point. The amplitude of an antinode is 2 × 10^–6 m. Write

(a) the equation of motion at a point 2 cm from the mid point and
(b) equation of the constituent waves in the rod. (Young’s Modulus of the material of the rod = 2 × 10¹¹ Nm^–2; density = 800 kgm^–3)

7. An aluminium wire of cross-sectional area 10^–6 m² is joined to a steel wire of the same cross-sectional area. This compound wire is stretched on a sonometer pulled by a weight of 10 kg. The total length of the compound wire between the bridges is 1.5 m of which the aluminium wire is 0.6 m and the rest is steel wire. Transverse vibrations are set up in the wire by using an external source of variable frequency. Find the lowest frequency of excitation for which the standing waves are formed such that the joint in the wire is a node. What is the total number of nodes at this frequency. The density of aluminium is 2.6 × 10³ kg/m³ and that of steel is 1.04 × 10⁴ kg/m³. (g = 10 m/s²)

8. A 3m long organ pipe open at both ends is driven to third harmonic standing wave. If the amplitude of pressure oscillations is 1 percent of mean atmospheric pressure (P0 = 10⁵ N/m²). Find the amplitude of particle displacement. Speed of sound v = 332 m/s and density of air ρ = 1.03 kg/m³

9. A boat is travelling in a river with a speed 10 m/sec along the stream flowing with a speed 2 m/sec. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible.

(a) What will be the frequency detected by a receiver kept inside the river downstream?
(b) Transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/sec in the direction opposite the river stream. Determine the frequency of the sound detected by the receiver. (Temperature of the air and water = 20° C; Density of river water = 10³ kg/m³; Bulk modulus of the water = 2.088 × 10⁹ Pa; Gas constant, R = 8.31 J/mol-K; Mean molecular mass of air = 28.8 × 10^–3 kg/mol; C_p / C_V for air = 1.4)

10. A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain heights in the pipe.

(a) Find the heights of water level (from the bottom of the pipe) at which resonances occur. Neglect end correction.
(b) Now the pipe is filled to a height H (≈3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H.
(c) If the radii of the pipe and the hole are 2 × 10^–2 m and 1 × 10^–3 m respectively, calculate the time interval between the occurrence of first two resonances. Speed of sound in air is 340 m/s and g = 10 m/s².

11. A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave pulse. Calculate:

(a) the time taken by the wave pulse to reach the other end R, and
(b) the amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint Q.

12. The air column in a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let P0 denote the mean pressure at any point in the pipe, and ΔP0 the maximum amplitude of pressure variation.

(a) Find the length L of the air column.
(b) What is the amplitude of pressure variation at the middle of the column?
(c) What are the maximum and minimum pressures at the open end of the pipe?
(d) What are the maximum and minimum pressures at the closed end of the pipe?

13. The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes. Speed of sound in air v = 330 m/s.

Advance Problems

1. 49%  2. πSρω²a²/4k    3. 95.5 s  4. (a) 33.33 m/s (b) 2250 Hz  5.f/√ 1+2at₀/c 6. (a) y = 10^–6 sin (0.1 π) sin (25000 πt) m (b) y1 = 10^–6 sin (25000 πt – 5 πx) m y2 = 10^–6 sin (25000 πt + 5 πx) m Here x is in metres and t is seconds. 7. 163.4 Hz, Five 8. 0.28 cm 9. (a) 1.0069 × 10⁵ Hz (b) 1.0304 × 10⁵ Hz 10. (a) 3.2 m, 2.4 m 1.6 m and 0.8 m (b) (-dH/dt)=(1.11× 10^-2)√H  (c) 43 s  11. (a) 0.14 s (b) 1.5 cm, 2.0 cm  12. (a) 15/16 m (b) ±ΔP0/√2 (c) P_max = P_min = P0(d) P_max = P0 + ΔP0, P_min = P0 – ΔP0 13. Length of closed organ pipe is l1 = 0.75 m while length of open pipe in either l2 = 0.99 m or 1.0067 m