If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced a distance 0.130 m from its equilibrium position and released with zero initial speed. Then after a time 0.803 s its displacement is found to be a distance 0.130 m on the opposite side, and it has passed the equilibrium position once during this interval.
A- Find the amplitude.
B- Find the period.
C- Find the frequency
2007-01-03 01:42:23 · 2 answers · asked by hadoi 1 in Science & Mathematics ➔ Physics
A=.130m
Amplitude is defined as a maximum displacement above the center of oscillations(equilibrium position).
y=Asin(wt) where w=2 pi f
and since there is no friction the oscillation has no damping and the block will move continuously with the same aplitude.
T=2(.803)s=1.606s (time from max to back to max point)
f=1/T=.623Hz
2007-01-03 05:41:04 · answer #1 · answered by Edward 7 · 4⤊ 0⤋
The amplitude is 0.260 m, the total movement since it can't go further than the initial displacement
The period is 1.606 s since it reached the same displacement but negative in 0.803 s, it is stopped and will return in the same time.
The frequency is 1/period.
2007-01-03 01:52:23 · answer #2 · answered by Mike1942f 7 · 1⤊ 1⤋