HARD
Physics
IMPORTANT
Earn 100

A wave is given by the equation, y=10sin2π100t-0.02x+10sin2π100t+0.02x. Find the loop length, frequency, velocity and maximum amplitude of the stationary wave produced.

Important Questions on Superposition and Standing Waves

HARD
Physics
IMPORTANT
A uniform horizontal rod of length 0.40 m and mass 1.2 kg is supported by two identical wires as shown in figure. Where should a mass of 4.8 kg be placed on the rod, so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone? g=10 m s-2

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HARD
Physics
IMPORTANT
An open organ pipe has a fundamental frequency of 300 Hz. The first overtone of a closed organ pipe has the same frequency as the first overtone of the open pipe. Find the length of each pipe. The velocity of sound in air =350 m s-1
HARD
Physics
IMPORTANT
An open organ pipe of length 11 cm in its fundamental mode vibrates in resonance with the first overtone of a closed organ pipe of length 13.2 cm filled with some gas. If the velocity of sound in the air is 330 m s-1, calculate the velocity of sound in the unknown gas.
MEDIUM
Physics
IMPORTANT

Find the fundamental frequency and the first four overtones of a 15 cm pipe, if the pipe is closed at one end, if the speed of sound, v=330 m s-1

MEDIUM
Physics
IMPORTANT

Find the fundamental frequency and the first four overtones of a 15 cm pipe, if the pipe is open at both ends.

MEDIUM
Physics
IMPORTANT

Find the fundamental frequency and the first four overtones of a 15 cm pipe. How many overtones may be heard by a person of normal hearing in each of the above cases? The velocity of sound in air =330 m s-1

HARD
Physics
IMPORTANT
 A glass tube of length 1.5 m is filled completely with water; the water can be drained out slowly at the bottom of the tube. Find the total number of resonance obtained, when a tuning fork of frequency 606 Hz is put at the upper open end of the tube, vsomad =340 m s-1.
MEDIUM
Physics
IMPORTANT

A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be a displacement node. It is found that when the frequency of the diaphragm is 2000 Hz, a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, the stationary wave pattern disappears,  but another stationary wave pattern reappears at a frequency of 1600 Hz. Calculate the speed of sound in air,