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lesya [120]
4 years ago
6

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

he capacitor has a charge of 5.4 µC and the current in the inductor is zero. (a) How long after t = 0 will the current in the circuit be maximum? (b) What will be this maximum current?
Physics
1 answer:
alexira [117]4 years ago
4 0

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

f = \dfrac{1}{2\pi\sqrt{LC}}

time period

T = \dfrac{1}{f}=2\pi\sqrt{LC}

after time T/4 current reach maximum

 t = \dfrac{T}{4}

 t = \dfrac{2\pi\sqrt{LC}}{4}

 t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}

        t = 8.2 x 10⁻⁴ s

        t = 0.82 ms

b) using law of conservation

  \dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2

  I^2 = \dfrac{CV^2}{L}

  I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}

  I =\sqrt{\dfrac{Q^2}{CL}}

  I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}

       I = 0.010 A

       I = 10 mA

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
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f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

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Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

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E=K=\frac{1}{2}mv^2

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E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

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5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

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