An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
1 answer:
Length of the pipe = 0.39 m
Third harmonic frequency = 1400 Hz
For the third harmonic:
Wavelength =
The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength
Distance from center = 0.25 × wavelength
Distance =
Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm
Distance =
Distance from the center to the nearest anti - node = 6.5 cm
Hence, the nearest distance to the anti - node from the center = 6.5 cm
So, option C is correct.
You might be interested in
Answer:
β = 114 db
Explanation:
The intensity of sound in decibles is
β = 10 log
in most cases Io is the hearing threshold 1 10-12 W / cm²
let's calculate the intensity of each instrument
I / I₀ = 10 (β / 10)
I = I₀ 10 (β / 10)
trumpet
I1 = 1 10⁻¹² 10 (94/10)
I1 = 2.51 10⁻³ / cm²
Thrombus
I2 = 1 10⁻¹² 10 (107/10)
I2 = 5.01 10-2 W / cm²
low
I3 =1 1-12 (113/10) W/cm²
I3 = 1,995 10-1 W / cm²
when we place the three instruments together their sounds reinforce
I_total = I₁ + I₂ + I₃
I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1
I_total = 0.00251 + 0.0501 + 0.1995
I_total = 0.25211 W / cm²
let's bring this amount to the SI system
β = 10 log (0.25211 / 1 10⁻¹²)
β = 114 db
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
