for a damped harmonic oscillation , the equation of motion is md^2x/dt^2+(gamma)dx/dt+kx=o with m=0.025kg,(gamma)=0.07kg/s and k=85N/m.
*calculate the period of motion
*number of oscillation in which its amplitude will become half of its initial value
*the number of oscillation in which its mechanical energy will drop to half of its intial value
*its relaxation time
*quality factor
To calculate the period of motion, we need to find the angular frequency (ω) first. The equation of motion for a damped harmonic oscillator can be written as:
md^2x/dt^2 + γdx/dt + kx = 0
Comparing this equation with the standard form:
md^2x/dt^2 + 2βdx/dt + ω^2x = 0
We can see that ω^2 = k/m and 2β = γ/m, which implies ω = sqrt(k/m) and β = γ/(2m). Substituting the given values, we get:
m = 0.025 kg
γ = 0.07 kg/s
k = 85 N/m
ω = sqrt(k/m) = sqrt(85/0.025) ≈ 58.31 rad/s
The period (T) of motion is given by T = 2π/ω. Substituting the value of ω, we get:
T = 2π/ω = 2π/58.31 ≈ 0.107 s (rounded to three decimal places)
To find the number of oscillations in which the amplitude becomes half of its initial value, we need to consider the damping factor β. The equation for the amplitude with damping is:
A(t) = A0 • exp(-βt)
Where A(t) is the amplitude at time t, A0 is the initial amplitude, and exp(-βt) is the damping factor.
Since we want to find the number of oscillations when the amplitude becomes half of its initial value, we need to solve the equation A(t) = A0/2 for t.
A0 • exp(-βt) = A0/2
exp(-βt) = 1/2
-βt = ln(1/2)
t = ln(2)/β
Substituting the value of β, we get:
t = ln(2)/(γ/(2m)) = ln(2)/(0.07/(2*0.025)) ≈ 9.231 s
To find the number of oscillations in this time, we divide the total time by the period:
Number of oscillations = t / T ≈ 9.231 / 0.107 ≈ 86.236
Therefore, approximately 86 oscillations are required for the amplitude to become half of its initial value.
To find the number of oscillations in which the mechanical energy drops to half, we use the equation for mechanical energy in a damped harmonic oscillator:
E(t) = (1/2)kA(t)^2 + (1/2)m(dx/dt)^2
We can calculate the ratio of initial energy E0 to the final energy at a given time t:
E(t) / E0 = (1/2)kA(t)^2 / (1/2)kA0^2
E(t) / E0 = A(t)^2 / A0^2
We want to find the number of oscillations when E(t) / E0 = 1/2. Let's call this time t1:
A(t1)^2 / A0^2 = 1/2
A(t1) / A0 = 1/sqrt(2)
We know that A(t) = A0 • exp(-βt). Substituting this in, we get:
A0 • exp(-βt1) / A0 = 1/sqrt(2)
exp(-βt1) = 1/sqrt(2)
Taking the natural logarithm of both sides:
-βt1 = ln(1/sqrt(2))
t1 = ln(sqrt(2))/β
Substituting the value of β, we get:
t1 = ln(sqrt(2))/(γ/(2m)) = ln(sqrt(2))/(0.07/(2*0.025)) ≈ 17.464 s
The number of oscillations required is:
Number of oscillations = t1 / T ≈ 17.464 / 0.107 ≈ 163.265
Therefore, approximately 163 oscillations are required for the mechanical energy to drop to half of its initial value.
The relaxation time for a damped harmonic oscillator is given by:
Relaxation time = 1 / β
Substituting the value of β, we get:
Relaxation time = 1 / (γ/(2m)) = 2m / γ = 2*0.025 / 0.07 ≈ 0.714 s (rounded to three decimal places)
Finally, the quality factor (Q) for a damped harmonic oscillator is given by:
Q = ωm / 2β
Substituting the values of ω and β, we get:
Q = (sqrt(k/m)) / (γ/(2m)) = 2m*sqrt(k/m) / γ = 2*0.025*sqrt(85/0.025) / 0.07 ≈ 62.857 (rounded to three decimal places)
To solve the given problem, we will use the equation of motion for a damped harmonic oscillator:
m(d^2x/dt^2) + γ(dx/dt) + kx = 0
1. Calculating the Period of Motion:
The period of motion (T) can be calculated using the angular frequency (ω), which can be found by solving the characteristic equation associated with the equation of motion:
mω^2 + γω + k = 0
To find the values of ω that satisfy this equation, we can use the quadratic formula:
ω = (-γ ± √(γ^2 - 4mk)) / (2m)
Substituting the given values, we get:
ω = (-0.07 ± √(0.07^2 - 4 * 0.025 * 85)) / (2 * 0.025)
Solving this equation will give us two possible values for ω, and the period can be calculated as:
T = 2π / ω
2. Number of Oscillations for Half Amplitude:
The amplitude of a damped harmonic oscillator decreases exponentially with time. Assuming the initial amplitude is A, we can determine the time it takes for the amplitude to become half (A/2) by using the exponential decay formula:
A(t) = A * e^(-t/τ)
Where τ is the relaxation time constant given by:
τ = 2m / γ
The number of oscillations required for the amplitude to become half is given by:
N = (ln 2) / (2π / ω)
3. Number of Oscillations for Half Mechanical Energy:
The mechanical energy (E) of a damped harmonic oscillator is given by:
E(t) = 0.5 * m * (dx/dt)^2 + 0.5 * k * x^2
Assuming the initial mechanical energy is E₀, the number of oscillations required for the mechanical energy to drop to half (E₀/2) can be calculated by:
N = (ln 2) / (2π / ω)
4. Relaxation Time:
The relaxation time (τ) was already calculated in step 2 as:
τ = 2m / γ
5. Quality Factor (Q):
The quality factor (Q) determines the sharpness of resonance and can be calculated as:
Q = ω₀ / (2 * β)
Where ω₀ is the natural frequency of the harmonic oscillator given by:
ω₀ = √(k / m)
And β is the damping coefficient defined by:
β = γ / (2m)
These steps will allow us to calculate the various parameters related to the given damped harmonic oscillation.