an object of mass 1.5 kg on a spring of force constant 600N/m loses 3% of its energy in each cycle.THE SYSTEM IS DRIVEn by a sinosuidal force with maximim value of F0=0.5N.(a) What is Q for this system?.(b) What is the resonance (angluar)frequency? (c) If the driving frequendy is varied what is the width delta w of the resonance? (d) What is the amplitude at resonance?(c) What is the amplutude if the driving frequency is w=19 rad\s?
Picture the Problem: We can find q from 1=2 PI(delta E\E)cycle and then ise this result to find the width of the resonance deltaw=w\Q.The resonance frequency is the natural frequency.The amplitude can be found from Equation 14-49 both at resonance and off resonance with damping contant caculated from Q using Equation 14-41 in the from b=w0m\Q
an object of mass 1.5 kg on a spring of force constant 600N/m loses 3% of its energy in each cycle.THE SYSTEM IS DRIVEn by a sinosuidal force with maximim value of F0=0.5N.(a) What is Q for this system?.(b) What is the resonance (angluar)frequency? (c) If the driving frequendy is varied what is the width delta w of the resonance? (d) What is the amplitude at resonance?(c) What is the amplutude if the driving frequency is w=19 rad\s?
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To solve this problem, we will follow these steps:
(a) To find Q for this system, we need to calculate the energy lost per cycle. The energy lost per cycle is given by δE, which is 3% of the total energy E. So, δE/E = 0.03.
Using the energy of a harmonic oscillator, E = (1/2)kA^2, where k is the force constant and A is the amplitude of oscillation.
Now, the angular frequency ω is related to the force constant k and the mass m by ω = sqrt(k/m). We can rewrite it as k = mω^2.
The total energy E can be written in terms of the mass m, angular frequency ω, and amplitude A as E = (1/2)kA^2 = (1/2)mω^2A^2.
Substituting this in the expression for δE/E, we get δE/E = (1/2)mω^2A^2 / E.
Now, substituting the given value of δE/E as 0.03 and rearranging the equation, we get Q = 1 / (2πδE/E) = 1 / (2π * 0.03).
Therefore, Q = 1 / (0.06π).
(b) The resonance (angular) frequency ω_r is the natural frequency of the system. This is given by ω_r = sqrt(k/m) = sqrt(600 N/m / 1.5 kg) = sqrt(400) rad/s.
(c) To find the width of the resonance Δω, we can use the formula Δω = ω_r / Q.
Substituting the values, Δω = sqrt(400) rad/s / (1 / (0.06π)) = sqrt(400) * (0.06π) rad/s.
(d) The amplitude at resonance can be found using the equation A_r = F_0 / (2π * m * ω_r * Q), where F_0 is the maximum value of the driving force.
Substituting the given value of F_0 as 0.5 N, we get A_r = 0.5 N / (2π * 1.5 kg * sqrt(400) rad/s * (1 / (0.06π))).
Simplifying this expression, we find A_r = 0.5 N / (2π * 1.5 kg * 20 * (0.06π)).
Lastly, to find the amplitude when the driving frequency is ω = 19 rad/s, we can use the equation A = F_0 / (2π * m * ω * Q).
Substituting the given values, we get A = 0.5 N / (2π * 1.5 kg * 19 rad/s * (1 / (0.06π))).