Answer:
a. 2.1 s
b.0.48 Hz
c. A=24cm
d. 72cm/s
Explanation:
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?
What are the period,
period is the time taken for a wave particle to make one complete oscillation
a) 31 / 15 = 2.066 seconds
= 2.1 s
(b) frequency
: this the number of oscillation made in one seconds.
it is also the inverse of the period.
= oscillations / time
= 15/31= 0.48 Hz
(c) amplitude
: maximum displacement from the origin
amplitude = 1/2 of the difference of oscillation marks
= 1/2(57-10) = 47/2cm
23.5cm
A=24cm
(d) maximum speed of the glider?
V=ωA
angular frequency *Amplitude
V=a*pi*f*amplitude
2π x frequency x amplitude = maximum speed
= 2π x .48 x 24
=72.38 cm/s
72cm/s
