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Chapter 10

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Chapter 10

Sinusoidally Driven Oscillations

- How do the characteristic frequencies generated in one object (say a piano string) excite vibrations in another object (say a sounding board)?

- If the board is a door, then the natural frequency is around 0.4 Hz.
- If the system is driven at 0.4 Hz, large amplitudes result.
- Smaller amplitudes result for driver frequency different from 0.4 Hz.

- The door starts with complex motions (transient) that settle down to sinusoidal, no matter the motor rate.
- The final frequency is always the driving frequency of the motor (w).
- The amplitude of the oscillations depends on how far from the natural frequency the motor is.

Natural Frequency

- Motor frequency is far below the natural frequency (w << wo) door moves almost in step with motor.
- Door moves toward motor when bands are stretched most.

- Door lags behind the motor.

- Door lags by one quarter cycle.

- Door lags by one-half cycle.

w/wo

Door Lags

<< 1

0

< 1

Small

= 1

¼-cycle

>> 1

½-cycle

Click on the link and experiment

- Transients are reproducible
- If crank starts in the same position, we get the same transient

- Damped Harmonic Oscillations
- Shown by changing the damping
- Imagine the bottom of the door immersed in an oil bath
- The amount of immersion gives the damping

- Damped harmonic oscillation (transient) is at the natural frequency
- Driven (steady state) oscillation is at the driver frequency

- As long as we are far from natural frequency, damping doesn’t affect the steady state.
- Near the natural frequency, damping does have an effect.

Small damping

W½

Amplitude

Large damping

Frequency

As damping is increased the height of the peak decreases

- As damping increases we expect the halving time to decrease ( )Oscillations die out quicker for larger damping.

- As damping increases the maximum amplitude decreases ( )

- Also notice W½ D.Larger damping means a broader curve.

- Range of frequencies for which the response is it least half the maximum amplitude.
- Let N be the number of oscillations that the pendulum makes in T½.
- Direct measurement yields
PBW = 38.2/N measured in %

- Imagine tuning an instrument by using a tuning fork (A 440) while playing A.
- If you are not matching pitch, the tuning fork is not being driven at its natural frequency and the amplitude will be small.
- Only at a frequency of 440 Hz will the amplitude of the tuning fork be large

- T½ = 5 sec (it takes about 5 seconds for the tuning fork to decay to half amplitude)
- N = (440 Hz) (5 sec) = 2200 cycles
- So when you get a good response from the tuning fork, you have found pitch to better than
PBW = 38.2/2200 = 0.017%

or 0.076 Hz!

- You must play long, sustained tones
- Short “toots” will stimulate the transient which recall is at the natural frequency of the tuning fork (440 Hz)
- Without the sustained driving force of the instrument, we will never get to the steady state and the tuning fork will ring due to the transient.
- You will think the instrument is in pitch when it is not.

- Each mode has its own frequency, decay time, and shape.
- The modes are always damped sinusoidal.
- Superposition applies.

Let Mode 1 have a natural frequency of 10 Hz and Mode 2 a natural frequency of 17.32 Hz.

17.32 Hz

10 Hz

Amplitude

Frequency

- Mass one has a mode one component and should lag a half-cycle behind the driver(w > wo1)
- Mass one also has a mode two component to its motion, and here the driving frequency is less than the natural frequency (w << wo2)
- Mass one keeps in step with the driver

- These conflicting tendencies account for the small amplitude here

- Driving Point Response Curve – measure the response at the mass being driven
- Transfer Point Response Curve – measure the response at another mass in the system (not a driven mass)

- At startup there is a transient that is made up of the damped sinusoids of all of the natural frequencies.
- Once the transient is gone the steady state is at the driving frequency. When the driving frequency is close to one of the natural frequencies, the amplitude is a maximum and resembles that natural mode.

The tray is clamped at three places. Sensors ( )and drivers ( )are used as pairs in the locations indicated.

- Sensor cannot pickup any mode whose nodal line runs through it.
- Notice that Sensor 2 is on the centerline
- It cannot pick up modes with nodal lines through the center, such as…

- If a driver falls on the nodal line of a mode, that mode will not be excited
- If a driver falls between nodal lines of a mode, that mode will be excited

- Superposition of all the modes excited and their amplitudes at the detector positions.
- Some modes may reinforce or cancel other modes.

- Example – consider the modes on the next screen
- Colored sections are deflected up at this time and the uncolored sections are deflected down
- The vertical lines show where in the pattern of each we are for a particular position on the plate

- Altering the location of either the driver or the detector will greatly alter what the transfer response curve will be.
- Altering the driver frequency will also change the response.

- Deflections of the same sign (giving a larger deflection)Add
- Deflections of opposite sign (canceling each other out)Subtract
- Deflection of one mode lined up with the node of the other (deflection due to one mode only)Single

196, 392, 588, 784, 980, 1176, …

Depress G3 slowly

Press & release G4

392, 784, 1176, 1568, 1960, 2352, …