In the oscillatory circuit, the voltage on the capacitor plates varies according to the law U=10 cos 104t V. The capacitance of the capacitor is 10 μF. It is necessary to find the inductance of the circuit and the law of change in current strength in it.
Answer:
Let's find the oscillation period: T=1/f=2π/104≈0.06 s.
Capacitance of the capacitor C = 10 µF.
Maximum capacitor voltage Umax=10 V.
Maximum loop current:
Imax= Umax/XL = Umax / (ωL) = Umax / (2πfL) = Umax / (2π×104×L)
The value of reactance XL is 1/(ωC).
Therefore, XL=1/(2πfC)=1/(2π×104×10×10-6) ≈ 1.5 Ohm.
Hence, L= XL/(2πf) ≈ 2.3 mH.
The law for changing the current in the circuit can also be found using the formula:
I=Imaxsin(ωt+Φ),
where Φ is the initial phase.
Then:
I = Imaxsin(ωt+Φ) = Umax/(ωL)sin(ωt+Φ) = Umax/(2πfL)sin(ωt+Φ).
Task 3.16
The equation for a plane wave propagating in an elastic medium is s = 10-8 sin (6280t- 1.256x:). It is necessary to determine the wavelength, the speed of its propagation and the oscillation frequency.
Answer:
The plane wave equation is:
s = A sin (kx - ωt + φ),
where A is the oscillation amplitude, k is the wave number, ω is the angular frequency, φ is the initial phase.
Comparing with the given equation, you can get:
A = 10-8 m;
k = 1.256 m-1;
ω = 6280 rad/s.
Wavelength λ is related to wavenumber k as follows:
λ = 2π/k ≈ 50 µm.
The speed of wave propagation is defined as v = λf, where f is the oscillation frequency:
v = λf = ω/k ≈ 4×104 m/s.
Oscillation frequency f = ω/2π ≈ 1×103 Hz
Task 3.26
Light with a wavelength of 0.72 μm is normally incident on a thin glass wedge. The distance between adjacent interference fringes in reflected light is 0.8 mm. The refractive index of glass is 1.5. It is necessary to determine the angle between the surfaces of the wedge.
Answer:
The distance between adjacent interference fringes in reflected light is related to the path difference between the rays reflected from the top and bottom surfaces of the wedge as follows:
2d = mλ,
where m is an integer (interference order).
Since the wedge is thin, we can assume that the angle of incidence of light on the wedge is small. Then the angle between the wedge surfaces θ is related to the refractive index of glass n as follows:
n = 1 + (d/λ)tanθ.
Substituting the value of the wavelength λ and the refractive index n, we obtain:
2d = mλ = 1.5λ,
whence d = 0.75λ ≈ 0.54 µm.
Substituting the value of d into the equation for the refractive index, we get:
n = 1 + (d/λ)tanθ = 1 + 0.75tanθ,
whence tanθ = (n-1)/0.75 ≈ 0.33.
Therefore, the angle between the wedge surfaces is θ ≈ 18.4°.
Task 3.36
Light falls normally onto a diffraction grating with a period of 6 µm. It is necessary to determine which spectral lines corresponding to wavelengths lying within the visible spectrum will coincide in the direction of an angle of 30°.
Answer:
A diffraction grating is a set of parallel slits with the same distance between them (period). When light passes through the grating, diffraction occurs, and an interference pattern can be observed on the screen in the form of a spectrum. For the nth-order maximum of the diffraction maximum, the following condition is satisfied:
dsinθ = nλ,
where d is the grating period, θ is the angle between the direction of the incident light and the normal to the grating, λ is the wavelength.
To determine which spectral lines will coincide in the direction of the 30° angle, it is necessary to consider the corresponding values of the angles θ and wavelengths λ for the visible spectrum. The visible spectrum lies in the wavelength range from 400 to 700 nm.
Substituting the values d=6 µm and θ=30° into the equation for the diffraction maximum, we obtain:
λ = dsinθ/n = 6×10-6×sin30°/n.
For n
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Solution 18.3.8 from the collection (workbook) of Kepe O.E. 1989 - В данной задаче необходимо определить модуль силы F,… ...