Respuesta :
Answer:
a) Time period is 3.3 seconds
b) The frequency is 0.3030 Hz
c) amplitude is 0.25 m
d) maximum speed is 0.476 m/s
Explanation:
Given the data in the question;
a) Period
Time Period T = Time taken for one oscillation
T = 33s / 10 = 3.3 seconds
Therefore, Time period is 3.3 seconds
b) Frequency
we know that frequency is the inverse of time period
so;
Frequency f = 1/T = 1 / 3.3 s
Frequency f = 0.3030 Hz
Therefore, The frequency is 0.3030 Hz
c) amplitude
amplitude A = [tex]\frac{1}{2}[/tex]( 60 cm - 10 cm )
A = [tex]\frac{1}{2}[/tex] × 50 cm
A = 25 cm
A = 0.25 m
Therefore, amplitude is 0.25 m
d) maximum speed of the glider
maximum speed [tex]V_{max}[/tex] = ωA
and ω = 2π/T
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{T}[/tex]A
so we substitute
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{3.3}[/tex] × 0.25 m
so maximum speed [tex]V_{max}[/tex] = 0.476 m/s
Therefore, maximum speed is 0.476 m/s
The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
Based on the given information,
• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.
• In 33 seconds, the glider completes 10 oscillations.
There is a need to find, the period, frequency, amplitude, and maximum speed of the glider.
a) Time period (T) for one oscillation is,
[tex]\frac{33s}{10} = 3.3 s[/tex]
b) The frequency (f) is the reciprocal of the time period,
[tex]f = \frac{1}{T} =\frac{1}{3.3S} = 0.303 Hz[/tex]
c) The amplitude (A) is,
[tex]A = \frac{1}{2} (60 cm-10cm) = 25cm[/tex]
[tex]A = 0.25 m[/tex]
d) Maximum speed of the glider is,
[tex]Vmax = \frac{2\pi }{T} (0.25 m)[/tex]
[tex]Vmax = 0.47575 m/s[/tex]
Thus, time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
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