Αποτελέσματα Αναζήτησης
We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let’s try one example of each. A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs.
(a) Calculate the frequency of the damped oscillation. (b) By what percentage does the amplitude of the oscillation decrease in each cycle? (c) Find the time interval that elapses while the energy of the system drops to 5.00% of its initial value. Solution (a) The natural angular frequency of the system ω is: ω = r k m = s 2.05×104 (N/m) 10. ...
measure the time period (T) of the oscillations by measuring the time taken by the pendulum to move from the equilibrium position, to the maximum displacement to the left, then to the maximum displacement to the right and back to the equilibrium position.
By rearranging the above formula so that its subject is frequency, you can derive the following formula for the time period of oscillations (T): T = 1 = 2π. f ω. Using the measurements described in the section above, you can use the following formulas with simple harmonic oscillators: = x Acos ωt. v = − A ωsin ωt. a = − A ω 2 cos ωt.
The period of a simple pendulum is T = 2\(\pi \sqrt{\frac{L}{g}}\), where L is the length of the string and g is the acceleration due to gravity. The period of a physical pendulum T = 2\(\pi \sqrt{\frac{I}{mgL}}\) can be found if the moment of inertia is known.
The period of the oscillator, and hence the frequency, can be determined by setting the oscillator (such as a pendulum or a mass on a spring) in to motion, and using a stopwatch to measure the time taken for one oscillation.
If you wish to see the sort of gyrations necessary to get a formula that can be solved with a calculator, see for example F.M.S. Lima and P. Arun, “An accurate formula for the period of a simple pendulum oscillating beyond the small angle regime”,
