Question

Очень срочно,помогите пожалуйста:1) Период электромагнитных колебаний увеличился в два раза.Собственная круговая частота1. не изменилась2. увеличилась в два раза3. уменьшилась в два раза4. увеличилась в раз 2) Индуктивность катушки колебательного контурауменьшилась в 9 раз. Собственная круговая частотаэлектромагнитных колебаний1. уменьшилась в три раза2. увеличилась в три раза3. уменьшилась в девять раз4. увеличилась в девять раз3) В момент когда заряд на обкладках конденсатораколебательного контура максимальна вся энергия контура1. равна энергии электрического поля конденсатора2. равна энергии магнитного поля катушки3. поровну разделена между энергией электрического поля иэнергией магнитного поля4. равна нулю

Answer 1

1). w = 2П/Т.  Если период увеличился в 2 раза, то w уменьшилась в 2 раза.

Ответ: 3. уменьшилась в 2 раза.

 

2) w = 1/кор(LC),  Если L уменьшиласт в 9 раз, то w - увеличилась в 3 раза.

Ответ: 2. увеличилась в 3 раза.

 

3) Когда заряд на конденсаторе максимален, вся энергия сосредоточена в нем и равна энергии электрического поля конденсатора.

Wэл = Q^2 /(2C).

Ответ: 1. равна энергии эл. поля конденсатора.

Answer 2

Period of Electromagnetic Oscillations

If the period of electromagnetic oscillations has doubled, we need to determine how this affects the natural angular frequency of the oscillations.

The natural angular frequency, denoted by ω, is related to the period T by the equation:

ω = 2π / T

where π is the mathematical constant pi.

Since the period has doubled, the new period T' is equal to 2T. Substituting this into the equation, we can find the new natural angular frequency ω':

ω' = 2π / T' = 2π / (2T) = π / T

Comparing this with the original natural angular frequency ω, we can see that the new natural angular frequency ω' is equal to half of the original natural angular frequency ω.

Therefore, the answer to the first question is:

2. The natural angular frequency has decreased by a factor of 2.

Inductance of the Oscillating Circuit

If the inductance of the oscillating circuit has decreased by a factor of 9, we need to determine how this affects the natural angular frequency of the electromagnetic oscillations.

The natural angular frequency ω of an oscillating circuit is related to the inductance L by the equation:

ω = 1 / √(LC)

where C is the capacitance of the circuit.

If the inductance L is decreased by a factor of 9, the new inductance L' is equal to L/9. Substituting this into the equation, we can find the new natural angular frequency ω':

ω' = 1 / √(L'C) = 1 / √((L/9)C) = 1 / (3√(LC))

Comparing this with the original natural angular frequency ω, we can see that the new natural angular frequency ω' is equal to one-third of the original natural angular frequency ω.

Therefore, the answer to the second question is:

3. The natural angular frequency has decreased by a factor of three.

Energy Distribution in the Oscillating Circuit

When the charge on the plates of the capacitor in the oscillating circuit is at its maximum, the energy in the circuit is distributed between the electric field energy of the capacitor and the magnetic field energy of the inductor.

The energy stored in the electric field of the capacitor is given by the equation:

E_electric = (1/2) * C * V^2

where C is the capacitance of the capacitor and V is the voltage across the capacitor.

The energy stored in the magnetic field of the inductor is given by the equation:

E_magnetic = (1/2) * L * I^2

where L is the inductance of the inductor and I is the current flowing through the inductor.

When the charge on the plates of the capacitor is at its maximum, the voltage across the capacitor is also at its maximum. This means that the energy stored in the electric field of the capacitor is at its maximum.

Since the energy in the circuit is conserved, the energy stored in the magnetic field of the inductor must be zero when the energy stored in the electric field of the capacitor is at its maximum.

Therefore, the answer to the third question is:

4. The energy of the magnetic field is zero when the charge on the plates of the capacitor is at its maximum.