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Chapter 10: Problem 69

Because of dissipative forces, the amplitude of an oscillator decreases $5.00 \%$ in 10 cycles. By what percentage does its energy decrease in ten cycles?

Short Answer

Expert verified
Answer: The approximate percentage decrease in the oscillator's energy is 9.79%.

Step by step solution

01

Understand the problem and relationship between amplitude and energy.

In this problem, we're dealing with an oscillator whose amplitude decreases due to dissipative forces. The energy of an oscillator is given by: \(E \propto A^2\) Where E is the energy, A is the amplitude, and the symbol \(\propto\) means "proportional to". The relationship between energy and amplitude is given by: \(E_1 / E_2 = (A_1 / A_2)^2\) Where \(E_1\) is the initial energy, \(E_2\) is the final energy, \(A_1\) is the initial amplitude, and \(A_2\) is the final amplitude after 10 cycles.
02

Calculate the final amplitude.

We're given that the amplitude decreases by 5% in 10 cycles. Let's first find the final amplitude as a fraction of the initial amplitude: \(A_2 = A_1 (1 - 0.05)\) => \(A_2 = 0.95 A_1\)
03

Calculate the ratio between the initial and final energies.

Now that we have the final amplitude expressed in terms of the initial amplitude, we can substitute this value into the energy ratio equation from Step 1: \(E_1 / E_2 = (A_1 / A_2)^2\) => \(E_1 / E_2 = (A_1 / (0.95A_1))^2\) First, we can simplify the equation by canceling out \(A_1\) in the denominator: \(E_1 / E_2 = (1 / 0.95)^2\) => \(E_1 / E_2 \approx 1.1081\) This value represents the ratio of initial energy to final energy after 10 cycles.
04

Calculate the percentage decrease in energy.

To find the percentage decrease in energy, we can subtract 1 from the ratio found in step 3 and multiply by 100: Percentage decrease = \((1 - (1 / 1.1081)) \times 100\) Percentage decrease \(\approx 9.79 \%\) Thus, the energy of the oscillator decreases by approximately \(9.79 \%\) in ten cycles due to dissipative forces.

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