Advertisements
Advertisements
प्रश्न
The displacement of a string is given by y (x, t) = 0.06 sin (2πx/3) cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10−2 kg.
- It represents a progressive wave of frequency 60 Hz.
- It represents a stationary wave of frequency 60 Hz.
- 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.
- Amplitude of this wave is constant.
उत्तर
b and c
Explanation:
We know that the standard equation of stationary wave is `y(x, t) = a sin(kx) cos(ωt)`
Given equation is `y(x, t) = 0.06 sin((2πx)/3) cos(120 πt)`
a. Comparing with a standard equation of stationary wave `y(x, t) = a sin(kx) cos(ωt)` Clearly, the given equation belongs to stationary wave. Hence, option (a) is not correct.
b. By comparing,
ω = 120 π
⇒ 2πf = 120 π
⇒ f = 60 Hz
c. k = `(2π)/3 = (2π)/λ`
⇒ λ = wavelength = 3m
Frequency = f = 60 Hz
Speed = v = fλ = (60 Hz) (3m) = 180 m/s
d. Since in stationary waves, all particles of the medium execute SHM with varying amplitude nodes.
APPEARS IN
संबंधित प्रश्न
Explain why (or how) In a sound wave, a displacement node is a pressure antinode and vice versa,
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.
Two wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the second overtone of second wire and radius of first wire is twice the radius of the second then the ratio of length of first wire to second wire is
The number of possible natural oscillations of the air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz? (v = 340 m/s)
Which of the following statements are true for a stationary wave?
- Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
- All the particles cross their mean position at the same time.
- All the particles are oscillating with same amplitude.
- There is no net transfer of energy across any plane.
- There are some particles which are always at rest.
An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?
The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its wavelength.
Two identical strings X and Z made of same material have tension Tx and Tz in them If their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio `"T"_x/"T"_"z"` is ______.
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm.
(Take the speed of sound in air as 340 ms-1.)