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Chapter 10: Problem 82

Show, using dimensional analysis, that the frequency \(f\) at which a mass- spring system oscillates radical is independent of the amplitude \(A\) and proportional to Whim. IHint: Start by assuming that \(f\) does depend on \(A\) (to some power).]

Short Answer

Expert verified
Answer: Yes, the frequency of a mass-spring system is independent of the amplitude. Our dimensional analysis has shown that the power of amplitude in the relationship between frequency, mass, spring constant, and amplitude is negative, which means that changes in amplitude will not influence the frequency directly.

Step by step solution

01

1. Identify the relevant quantities

For a mass-spring system, the relevant quantities are mass (\(M\)), spring constant (\(k\)), and the amplitude (\(A\)). Additionally, the frequency (\(f\)) is the property we want to find a relationship for.
02

2. Write the dimensions of the relevant quantities

Now, we need to express each of these quantities in terms of the base dimensions of mass (\(M\)), length (\(L\)), and time (\(T\)). - Mass: \([M] = M^1L^0T^0\) - Spring constant: \([k] = \frac{ML}{T^2}\), since it is a force per unit of length - Amplitude: \([A] = L^1\), as it is a length - Frequency: \([f] = T^{-1}\), because it is the number of oscillations per unit of time
03

3. Assume a relationship between the frequency and the other quantities

Suppose that there is a relationship between the frequency \(f\), mass \(M\), spring constant \(k\), and amplitude \(A\) of the form: \(f = C M^a k^b A^c\), where \(a\), \(b\), and \(c\) are the powers to which the quantities are taken, and \(C\) is a constant. Now, we apply dimensional analysis to this equation.
04

4. Equate dimensions on both sides of the equation

Using the dimensions of the quantities found in step 2, we write: \([f] = [C M^a k^b A^c]\) \(T^{-1} = M^{a} \left(\frac{ML}{T^2}\right)^b L^c\) \(T^{-1} = M^{a + b}L^{b + c}T^{-2b}\) Now, for the equation to hold true, the powers of \(M\), \(L\), and \(T\) must be the same on both sides of the equation.
05

5. Compare dimensions and deduce the powers

Comparing the powers of the base dimensions, we get: For mass: \(a + b = 0 \Rightarrow a = -b\) For length: \(b + c = 0 \Rightarrow c = -b\) For time: \(-2b = -1 \Rightarrow b = \frac{1}{2}\) Now, we can find the values of \(a\) and \(c\) using the values of \(b\), and we get: \(a = -\frac{1}{2}\) and \(c = -\frac{1}{2}\)
06

6. Interpret the result

The relationship we found is: \(f = C M^{-\frac{1}{2}} k^{\frac{1}{2}} A^{-\frac{1}{2}}\) Notice that the power of the amplitude \(A\) is negative, so if we change \(A\), it will not influence the frequency directly. Thus, the frequency of the mass-spring system is independent of the amplitude.

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Most popular questions from this chapter

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