Only if they move opposite to each other can they superpose in such a way that a standing wave is formed.
CloseThese are the points on the wave when the 2 superposing waves cancel each other out.
CloseSince the particles are only moving up and down, they cannot transfer any energy in a progressive way.
CloseAn object is critically damped if it returns to the equilibrium very quickly, but without ever passing the equilibrium point.
CloseOn 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.
CloseStanding waves result from the interference of two identical waves traveling in opposite directions, leading to points of no displacement (nodes) and maximum displacement (antinodes).
CloseThe distance between two adjacent nodes (or two adjacent antinodes) is half the wavelength: \[ \lambda = 2 \times \text{(distance between nodes)} = 2 \times 0.5 = 1.0 \, \mathrm{m}. \]
CloseThe 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: \[ \Delta \phi = 90^\circ. \]
CloseFor 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: \[ f = \frac{v}{\lambda} = \frac{340}{3.4} = 100 \, \mathrm{Hz}. \]
CloseAt resonance, the driving frequency matches the natural frequency of the system, resulting in maximum energy transfer and maximum amplitude.
CloseDamping reduces the amplitude of oscillations at resonance and slightly lowers the resonant frequency because the system loses energy more rapidly.
CloseResonance 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).
CloseFor 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: \[ f = \frac{v}{\lambda} = \frac{240}{1.2} = 200 \, \mathrm{Hz}. \]
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