A damped harmonic oscillator consists of a block (m = 2.00 kg), a spring (k = 10.0 N/m), and a damping force (F = −bv). Initially, it oscillates with an amplitude of 25.0 cm; because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. We assume that the damping force does not affect the period of each oscillation. (a) What is the value of b? (b) Justify the assumption that "damping force does not affect the period of each oscillation".

Respuesta :

Explanation:

Given that,

Mass of block = 2.00 kg

Spring constant = 10.0 N/m

Initial amplitude [tex]x_{i}= 25.0[/tex]

Final amplitude [tex]x_{f}= 0.75 x_{i}[/tex]

Number of oscillation = 4

(a). We need to calculate the time

Given formula,

[tex]t = nT[/tex]

[tex]t=n2\pi\sqrt{\dfrac{m}{k}}[/tex]

Put the value into the formula

[tex]t=4\times2\pi\sqrt{\dfrac{2.00}{10.0}}[/tex]

[tex]t=11.23\ sec[/tex]

We need to calculate the value of b

Using damping formula

[tex]x_{m}e^{\dfrac{-bt}{2m}}=0.75x_{i}[/tex]

[tex]e^{\dfrac{-b\times11.23}{2\times2.00}}=0.75[/tex]

[tex]e^{-2.81b}=0.75[/tex]

[tex]-2.81 b=ln(0.75)[/tex]

[tex]b=\dfrac{0.28}{2.81}[/tex]

[tex]b=0.099\ kg/s[/tex]

(b), The damping force equation is

[tex]m\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+kx=0[/tex]

We know that,

The time period is

[tex]T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}+(\dfrac{b}{2m})^2}}[/tex]

Therefore, time period is affected by damping force.

Hence, This is the required solution.

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