Pressure Wave Derivation
Trending Questions
- 1000π Hz, 500 m/s
- 200π Hz, 500 m/s
- 1500π Hz, 10003 m/s
- 100π Hz, 250 m/s
If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased ?
- A high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.
- A high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open
- A low-prressure pulse starts traveling up the pipe, if the other end of the pipe is open
- A low-pressure pulse stats traveling up the pipe, if the other end of the pipe is closed
Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm travelling in air. The pressure at a point varies between (1.0 × 105 and 14) Pa and the particles of the air vibrate in simple harmonic motion of amplitude 5.5× 10−6 m.
1.4 x 105 N/m2
3.2 x 102 N/m2
5.5 x 106 N/m2
None of these
- P
- Q
- R
- S
For sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2×10−8 m produces a pressure amplitude of 3.0×10−2 Pa. What is the wavelength of these waves? (Provided Bulk modulus of Air = 1.42× 105Pa)
0.4 m
0.3 m
0.36 m
None of these
(ρo is the density of the medium at pressure Po)
- ρ=ρosin(ωt−kx)
- ρ=−ρosin(ωt−kx)
- ρ=ρocos(ωt−kx)
- ρ=−ρocos(ωt−kx)
- ΔPosin(kx−ωt)
- ΔPocos(kx−ωt)
- −ΔPocos(kx−ωt)
- −ΔPocos(kx+ωt)
The equation of a sound wave in air is given by
p=(0.01 N m−2) sin [(1000s−1) t−(3.0 m−1)x]
(a) Find the frequency, wavelength and the speed of sound wave in air.
(b) If the equilibrium pressure of air is 1.0×105 N m−2, what are the maximum and minimum pressures at a point as the wave passes through that point.
(i) 2.1(p)frequency in Hz(ii)1.0×105+0.01 (q)wavelength in m(iii)333(r) speed in m/s(iv)160(s)Max p in Nm−2(p) - (iv); (q) - (i); (r) - (iii); (s) - (ii); (t) - (v)
(p) - (i); (q) - (ii); (r) - (iii); (s) - (iv); (t) - (v)
(p) - (iv); (q) - (v); (r) - (iii); (s) - (ii); (t) - (i)
(p) - (v); (q) - (ii); (r) - (i); (s) - (iii); (t) - (iv)
- 225
- 250
- 275
- 300
- 105 N/m2
- 4×105 N/m2
- 2×105 N/m2
- 3×105 N/m2
- 1.25×10−6 W/m2
- 1.56×10−6 W/m2
- 10−6 W/m2
- 2×10−6 W/m2
- ε0E20
- 12ε0E20
- 2ε0E20
- 14ε0E20
- ΔPosin(kx−ωt)
- ΔPocos(kx−ωt)
- −ΔPocos(kx−ωt)
- −ΔPocos(kx+ωt)
- 0.460 Pa
- 10Pa
- 50 Pa
- 1 Pa
- 2.2×10−8 m
- 2.2×10−9 m
- 2.2×10−10 m
- 2.2×10−11 m
Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm travelling in air. The pressure at a point varies between (1.0×105±14) Pa and the particles of the air vibrate in simple harmonic motion of amplitude 5.5×10−6m.
- 30 cm
- 40 cm
- 10 cm
- 20 cm
- beats will not be heard
- 6
- 12
- 34
Equation of wave is
- y=10−4sinπ(400t−0.8x)
- y=10−4sinπ(2000t−0.8x)
- y=10−4sinπ(100t−8x)
- y=10−4sinπ(100t−2x)
For sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2×10−8 m produces a pressure amplitude of 3.0×10−2 Pa. What is the wavelength of these waves? (Provided Bulk modulus of Air = 1.42× 105Pa)
0.3 m
0.4 m
0.36 m
None of these
- 1 : 2
- 2 : 1
- 4 : 5
- 5 : 3
- 5:6
- 6:7
- 5:9
- 5:4
(ρo is the density of the medium at pressure Po)
- ρ=ρosin(ωt−kx)
- ρ=−ρosin(ωt−kx)
- ρ=ρocos(ωt−kx)
- ρ=−ρocos(ωt−kx)
- 4:25
- 64:225
- 9:100
- 8:15
ΔP=12sin(8.18x−2700t+π/4)N/m2
Find the displacement amplitude. Density of air = 1.29kg/m3
Give answer in terms of 10−5 m
- 2
- 1.05
- 5
- 15
- 10−6 W/m2
- 1.25×10−6 W/m2
- 1.56×10−6 W/m2
- 2×10−6 W/m2