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A body of mass 10 kg vibrates on a vertical spring whose constant 2∙104 N/m. Air resistance can be expressed as a force which is proportional to the speed, and the constant b is 3 Ns/m. a) What is the frequency of damped vibration? b) By what percentage is the oscillation amplitude reduced during a period? c) How long does it take for the energy of the system to decrease to 5 % of the initial value?

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To answer the given questions, we need to use the equations and concepts related to damped vibrations. The differential equation that describes the motion of a damped harmonic oscillator is:

m * x'' + b * x' + k * x = 0

where:

m is the mass of the body (10 kg)

x'' is the acceleration of the body (second derivative of displacement with respect to time)

b is the damping constant (3 Ns/m)

x' is the velocity of the body (first derivative of displacement with respect to time)

k is the spring constant (2∙10^4 N/m)

x is the displacement of the body from its equilibrium position

a) To find the frequency of damped vibration, we first need to calculate the angular frequency (ω) using the formula:

ω = sqrt(k / m)

Substituting the given values:

ω = sqrt(2∙10^4 N/m / 10 kg) = sqrt(2000 rad/s)

The frequency (f) can be calculated using the formula:

f = ω / (2π)

Substituting the value of ω:

f = sqrt(2000 rad/s) / (2π) ≈ 25.23 Hz

Therefore, the frequency of damped vibration is approximately 25.23 Hz.

b) The percentage reduction in the oscillation amplitude during a period can be calculated using the formula:

Percentage reduction = (b / (2 * sqrt(k * m))) * 100

Substituting the given values:

Percentage reduction = (3 Ns/m / (2 * sqrt(2∙10^4 N/m * 10 kg))) * 100 ≈ 3.54%

Therefore, the oscillation amplitude is reduced by approximately 3.54% during a period.

c) The time taken for the energy of the system to decrease to 5% of the initial value can be determined using the equation for damped vibrations:

E(t) = E0 * exp(-2bt/m)

where:

E(t) is the energy at time t

E0 is the initial energy of the system

We want to find the time (t) when E(t) = 0.05E0.

0.05E0 = E0 * exp(-2bt/m)

Taking the natural logarithm (ln) of both sides:

ln(0.05) = -2bt/m

Solving for t:

t = (ln(0.05) * m) / (-2b)

Substituting the given values:

t = (ln(0.05) * 10 kg) / (-2 * 3 Ns/m) ≈ 0.96 seconds

Therefore, it takes approximately 0.96 seconds for the energy of the system to decrease to 5% of the initial value.

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