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Questions on forced vibrations and resonance
Resonance is the rise in amplitude that occurs when the driving frequency of a forced vibration matches the natural frequency of the system. Heavier damping lowers and broadens the peak and shifts it below the natural frequency, which is how unwanted resonance is engineered out. At the natural frequency the oscillator lags the driver by a quarter of a cycle.
18 original questions · 54 marks · the forced vibrations and resonance notes · Periodic motion
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Distinguish between free vibrations and forced vibrations.
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Free vibrations occur when a system is displaced and released, oscillating at its own natural frequency with no driving force (1). Forced vibrations occur when a periodic external driving force makes the system oscillate at the driving frequency (1).Define resonance.
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Resonance is the large-amplitude response of a system when it is driven at (or very close to) its natural frequency (1), so that energy is transferred to it most efficiently (1).State the condition for resonance to occur.
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The driving frequency must equal the natural frequency of the system (1), or be very close to it (1).Several pendulums of different lengths hang from the same taut horizontal cord. A driver pendulum of length 0.60 m is set swinging. State which of the other pendulums oscillates with the largest amplitude, and explain why.
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The pendulum whose length is also 0.60 m (1). Its natural frequency equals the driving frequency of the driver pendulum, so it resonates (1).A tuning fork marked 512 Hz is struck and held near a second, identical, unstruck fork. The second fork begins to sound. Explain why.
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The sound waves from the first fork act as a periodic driving force on the second fork at 512 Hz (1). This equals the natural frequency of the identical second fork, so it resonates and oscillates with large amplitude (1).A mass–spring system has a spring constant of 100 N m−1 and a mass of 0.25 kg. Calculate its natural frequency, and state the driving frequency that would cause resonance.
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f = (1/2π)√(k/m) (1)
= (1/2π)√(100/0.25) = 3.18 Hz (1)
Resonance occurs when the driving frequency equals this, about 3.18 Hz (1)Sketch and describe how the amplitude of a driven system varies with driving frequency.
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With light damping the graph rises to a sharp peak at the natural frequency and falls away on either side (1). At low frequencies the amplitude is small (1); it grows to a maximum at resonance, then decreases again at higher frequencies (1).A simple pendulum of length 0.50 m is used as a driven oscillator (g = 9.81 m s−2). Calculate its natural frequency, and state the driving frequency needed for resonance.
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T = 2π√(l/g) = 2π√(0.50/9.81) = 1.42 s (1)
f = 1/T = 0.70 Hz (1)
Resonance occurs when driven at about 0.70 Hz (1)Describe how increasing the amount of damping changes the resonance peak.
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More damping lowers the resonance peak, giving a smaller maximum amplitude (1). It also broadens the peak over a wider range of frequencies (1) and shifts it to a slightly lower frequency (1).The wing mirror of a van vibrates violently when the engine runs at 1500 revolutions per minute, but is nearly steady at other engine speeds. Determine the natural frequency of the mirror on its mounting, and explain what happens to the vibration when the engine speed rises to 3000 revolutions per minute.
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The engine provides a periodic driving force at 1500/60 = 25 Hz (1)
Violent vibration means resonance, so the natural frequency of the mirror is 25 Hz (1)
At 3000 rpm the driving frequency (50 Hz) is well above the natural frequency, so the amplitude is small and the mirror is nearly steady (1)A string is stretched between two fixed points 0.80 m apart. Waves travel along the string at 40 m s−1. Calculate the fundamental frequency of the string, and state what is observed when a vibration generator drives the string at this frequency.
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For the fundamental, λ = 2L = 1.6 m (1)
f = v/λ = 40/1.6 = 25 Hz (1)
A large-amplitude stationary wave forms on the string: the string resonates (1)A student drives a mass-spring system at a range of frequencies and records the amplitude. Driving frequency/Hz: 2.0, 2.25, 2.5, 2.75, 3.0. Amplitude/mm: 4, 9, 23, 8, 5. State the approximate natural frequency of the system, and explain why the amplitudes at 2.0 Hz and 3.0 Hz are much smaller than the maximum.
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The natural frequency is about 2.5 Hz, where the amplitude peaks (1). At 2.0 Hz and 3.0 Hz the driving frequency does not match the natural frequency, so the pushes fall out of step with the motion (1). Energy is then transferred to the system far less efficiently, and some pushes partly cancel the motion, so the amplitude stays small (1).A system has a natural frequency of 50 Hz. Describe what happens to its amplitude as the driving frequency is slowly increased from well below 50 Hz to well above it.
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Well below 50 Hz the amplitude is small (1). As the driving frequency approaches 50 Hz the amplitude grows rapidly, reaching a maximum at resonance (1). With light damping that maximum sits at 50 Hz; heavier damping moves it a little below 50 Hz and flattens it (1). As the frequency rises further above 50 Hz the amplitude falls again, becoming small at high frequencies (1).Explain why resonance can be dangerous for a structure such as a bridge, and state one way engineers reduce the risk.
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If a periodic force (such as wind or marching feet) matches the structure's natural frequency (1), resonance builds up large-amplitude oscillations (1) that can cause damage or collapse (1). Engineers add damping (for example tuned mass dampers) to absorb energy and limit the amplitude, or design the structure so its natural frequency avoids likely driving frequencies (1).Give one useful example and one problematic example of resonance.
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Useful: the tuning circuit of a radio resonating at the chosen station's frequency (or MRI, or a pushed swing) (1). Problematic: a bridge or building oscillating in the wind, or a washing machine vibrating badly at certain spin speeds (1).A car bounces on its suspension with a natural frequency of 1.3 Hz. It is driven along a road with regularly spaced ridges 12 m apart, where the speed limit is 13.4 m s−1. Deduce whether the car can experience resonance without breaking the speed limit.
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Each ridge gives the suspension one push, so the driving frequency is v/d, where d = 12 m (1)
Resonance needs v/d = 1.3 Hz, so v = 1.3 × 12 = 15.6 m s−1 (1)
15.6 m s−1 is above the 13.4 m s−1 limit (1)
So within the speed limit the driving frequency stays below the natural frequency and resonance cannot occur (1)Describe how the phase of a driven oscillator, relative to the driver, changes as the driving frequency increases from well below the natural frequency to well above it.
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Well below the natural frequency the oscillator moves in phase with the driver (1). At the natural frequency the oscillator lags the driver by a quarter of a cycle (90°), and by exactly that whatever the damping (1). Well above the natural frequency the oscillator is in antiphase with the driver, lagging by half a cycle (180°) (1).A thin metal panel on a machine rattles loudly, but only when the machine's motor runs within a narrow band of speeds. Explain fully why this happens, and describe two changes the designer could make to reduce the rattling. Use the ideas of forced vibrations, resonance and damping in your answer.
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The panel is an oscillator with its own natural frequency (1). The rotating motor applies a periodic driving force to the panel at the rotation frequency, so the panel performs forced vibrations (1). When the motor speed brings the driving frequency close to the panel's natural frequency, the panel resonates (1). At resonance energy is transferred from the motor to the panel at the maximum rate, so the amplitude becomes large and the panel rattles; away from this band the amplitude is small (1). Adding damping, for example a rubber or self-adhesive damping layer on the panel, lowers and broadens the resonance peak so the maximum amplitude is much smaller (1). Alternatively, stiffening the panel or adding mass changes its natural frequency so that it no longer lies within the motor's running range (1).
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