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A lightly damped harmonic oscillator losses energy at the rate of 4% per minute. The decrease in amplitude of the oscillator per minute will be closest to
- a)8 %
- b)6 %
- c)4 %
- d)2 %
Correct answer is option 'D'. Can you explain this answer?
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The given problem involves a lightly damped harmonic oscillator that loses energy at a rate of 4% per minute. We are required to determine the decrease in amplitude of the oscillator per minute.
To solve this problem, we can use the formula for the amplitude of a damped harmonic oscillator:
A(t) = A₀e^(-bt),
where A(t) is the amplitude at time t, A₀ is the initial amplitude, e is the base of the natural logarithm, b is the damping constant, and t is the time.
In this case, the oscillator loses energy at a rate of 4% per minute, which means that after one minute, the amplitude decreases by 4%. We can express this as:
A(1) = A₀ - 0.04A₀ = 0.96A₀.
Comparing this with the equation for the amplitude of a damped harmonic oscillator, we can see that the damping constant b is equal to 0.04. Plugging this value into the equation, we get:
A(t) = A₀e^(-0.04t).
Now, we need to determine the decrease in amplitude per minute. To do this, we can differentiate the amplitude equation with respect to time:
dA(t)/dt = -0.04A₀e^(-0.04t).
This equation gives us the rate of change of the amplitude with respect to time. To find the decrease in amplitude per minute, we substitute t = 1 into the equation:
dA(1)/dt = -0.04A₀e^(-0.04) = -0.04A₀(0.961) ≈ -0.0384A₀.
Now, we can calculate the percentage decrease in amplitude per minute:
Percentage decrease = (|dA(1)/dt| / A₀) * 100.
Substituting the values, we get:
Percentage decrease = (0.0384A₀ / A₀) * 100 = 3.84%.
Rounding this to the nearest whole number, we find that the decrease in amplitude per minute is closest to 4%, which corresponds to option (d).
To solve this problem, we can use the formula for the amplitude of a damped harmonic oscillator:
A(t) = A₀e^(-bt),
where A(t) is the amplitude at time t, A₀ is the initial amplitude, e is the base of the natural logarithm, b is the damping constant, and t is the time.
In this case, the oscillator loses energy at a rate of 4% per minute, which means that after one minute, the amplitude decreases by 4%. We can express this as:
A(1) = A₀ - 0.04A₀ = 0.96A₀.
Comparing this with the equation for the amplitude of a damped harmonic oscillator, we can see that the damping constant b is equal to 0.04. Plugging this value into the equation, we get:
A(t) = A₀e^(-0.04t).
Now, we need to determine the decrease in amplitude per minute. To do this, we can differentiate the amplitude equation with respect to time:
dA(t)/dt = -0.04A₀e^(-0.04t).
This equation gives us the rate of change of the amplitude with respect to time. To find the decrease in amplitude per minute, we substitute t = 1 into the equation:
dA(1)/dt = -0.04A₀e^(-0.04) = -0.04A₀(0.961) ≈ -0.0384A₀.
Now, we can calculate the percentage decrease in amplitude per minute:
Percentage decrease = (|dA(1)/dt| / A₀) * 100.
Substituting the values, we get:
Percentage decrease = (0.0384A₀ / A₀) * 100 = 3.84%.
Rounding this to the nearest whole number, we find that the decrease in amplitude per minute is closest to 4%, which corresponds to option (d).
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A lightly damped harmonic oscillator losses energy at the rate of 4% per minute.The decrease in amplitude of the oscillator per minute will be closest toa)8 %b)6 %c)4 %d)2 %Correct answer is option 'D'. Can you explain this answer?
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