Superpositions Flashcards
Progressive wave vs Stationary wave - 6
(Property:
Progressives wave;
Stationary wave)
Waveform:
Propagates with the vel of the wave;
Does not propagate
Energy:
Transports energy;
Does not transport energy
Amplitude:
Every pt oscillates with the same amplitude;
amplitude varies from 0 at the nodes to the max at the antinodes
Phase:
All particles within one wavelength have different phases;
All particles b/w two adj nodes have the same phase. Particles in adj segments have a phase difference of pi rad
Frequency:
All pts vibrate in s.h.m. with the freq of the wave;
Except for the nodes which are at rest, all pts vibrate in s.h.m. with the same frequency as the progressive waves that give rise to it
Wavelength:
Equals to the dist b/w adj pts which have the same phase;
Equals to twice the dist b/w a pair of adj nodes or antinodes
For stationary waves on strings:
Modes of vibration | Graphical rep. | λ | f | Aka…
Modes of vibration | Graphical rep. | λ | f | Aka…
Fundamental mode | 1 | L = 1 (λ/2) | f=v/2L | 1st harmonic
1st overtone | 2 antinodes…| L = 2 (λ/2) | f=2(v/2L)| 2nd harmonic
2st overtone | 3 antinodes…| L = 3 (λ/2) | f=3(v/2L)| 3rd harmonic
(n-1)th overtone | L = n (λ/2) | f=n(v/2L) | nth harmonic
Why must there be nodes at the fixed ends/pulley
A node is a pt on the stationary wave where the particle is always at rest. A node must exist at the pulley as the pulley. and hence the string, is fixed in position
Why is it necessary to adjust either the length of the string or the freq of the oscillator in order to obtain observable stationary waves on the string
Since the tension of the string is constant, the speed of the wave on the string is constant. Stationary waves will only be formed if the length of the string is equal to certain multiples of half-wavelength of the wave (i.e. when resonance occurs)
L = n X 1/2 λ n = 1,2,3
We can achieve this condition for a fixed freq f by adjusting the length L, or for a fixed length L by adjusting the frequency f
For stationary waves in closed pipes (1 side):
Modes of vibration | Graphical rep. | λ | f | Aka…
Modes of vibration | Graphical rep. | λ | f | Aka…
Fundamental mode | 1/2 | L = 1 (λ/4) | f=v/4L | 1st harmonic
1st overtone | 1.5 antinodes… | L = 3 (λ/4) | f=3 (v/4L) | 3rd harmonic
2st overtone | 2.5 | L = 5(λ/4) | f=5(v/4L) | 5th harmonic
(N-1)th overtone | L = (2n-1)(λ/4) | f=(2n-1)(v/4L) | (2n-1)th harmonic
For stationary waves in open pipes:
Modes of vibration | Graphical rep. | λ | f | Aka…
Modes of vibration | λ | f | Aka…
Fundamental mode | L = 1 (λ/2) | f=v/2L | 1st harmonic
1st overtone | L = 2 (λ/2) | f=2(v/2L)| 2nd harmonic
2st overtone | L = 3 (λ/2) | f=3(v/2L)| 3rd harmonic
(n-1)th overtone | L = n (λ/2) | f=n(v/2L) | nth harmonic
End correction
For closed pipe:
Wavelength:
L +c = 1 (λ/4) | f=v/4 (L+c)
For open pipes:
L+c = 1 (λ/2) | f=v/2(L+c)
Using cathode-ray oscilloscope
Since the distance L b/w 2 successive pressure antinodes (or displacement nodes) is L=1/2λ and the frequency f can be determined from the CRO, the speed of the sound can be calculated using
v=fλ = 2fL
Using resonance tube
Fundamental mode diagram
L1 + c = 1/4 λ — (a)
1st overtone diagram
L2 + c = 3/4 λ — (b)
As shows in the diagram, eqn (b) - (a)
1/2 λ = L2 - L1
Since the freq f of the tuning fork is known, speed of speed in air can be calc using
v = fλ = 2f (L2-L1)
Why can sound waves and water waves go round corners but light waves seem to travel only in straight lines?
OR can u hear but not see qns
The ratio of the wavelength to the size of common corners is large (ratio>1) for sound n water waves and small (ratio«1) for light waves. Hence, more sig diffraction is observed for sound and water waves than light.
