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Only if they move opposite to each other can they superpose in such a way that a standing wave is formed.

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These are the points on the wave when the 2 superposing waves cancel each other out.

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Since the particles are only moving up and down, they cannot transfer any energy in a progressive way.

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An object is critically damped if it returns to the equilibrium very quickly, but without ever passing the equilibrium point.

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Resonance is when the driving frequency is equal to the natural frequency.

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On an amplitude-frequency curve, introducing damping results in the peak of the curve being lower, and pushed to the left, to a smaller frequency, hence C.

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Standing waves result from the interference of two identical waves traveling in opposite directions, leading to points of no displacement (nodes) and maximum displacement (antinodes).

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The distance between two adjacent nodes (or two adjacent antinodes) is half the wavelength: $$\lambda = 2 \times \text{(distance between nodes)} = 2 \times 0.5 = 1.0 \, \mathrm{m}.$$

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The phase difference between points on a standing wave depends on their separation relative to the wavelength. A separation of $\frac{\lambda}{4}$ corresponds to a phase difference of: $$\Delta \phi = 90^\circ.$$

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For a pipe closed at one end, the fundamental wavelength is given by: $$\lambda = 4L.$$ $$\lambda = 4 \times 0.85 = 3.4 \, \mathrm{m}.$$ The fundamental frequency is: $$f = \frac{v}{\lambda} = \frac{340}{3.4} = 100 \, \mathrm{Hz}.$$

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At resonance, the driving frequency matches the natural frequency of the system, resulting in maximum energy transfer and maximum amplitude.

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Damping reduces the amplitude of oscillations at resonance and slightly lowers the resonant frequency because the system loses energy more rapidly.

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Resonance occurs when the driver frequency matches the natural frequency of the system. At resonance ($5.0 \, \mathrm{Hz}$), the amplitude of oscillation is maximized due to the efficient transfer of energy from the driver to the system. When the driver frequency is off-resonance ($4.5 \, \mathrm{Hz}$), the amplitude is significantly lower.

Therefore, the amplitude is greater when the driver frequency is equal to the natural frequency (5.0 Hz).

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For a string fixed at both ends, the wavelength of the second harmonic is: $$\lambda = \frac{2L}{n} \quad (n = 2),$$ $$\lambda = \frac{2 \times 1.2}{2} = 1.2 \, \mathrm{m}.$$ The frequency is given by: $$f = \frac{v}{\lambda} = \frac{240}{1.2} = 200 \, \mathrm{Hz}.$$