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amandanunnery33
21.12.2019 •
Physics
As your jet plane speeds down the runway on takeoff, you measure its acceleration by suspending your yo-yo as a simple pendulum and noting that when the bob (of mass 725 g) is at rest relative to you, the string (of length 52.8 cm) makes an angle of 33° with the vertical. the acceleration of gravity is 9.81 m/s^2. find the period for small oscillations of this pendulum.
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Ответ:
T = 1.46 s
Explanation:
The time period of pendulum can be given as:
Where,
L = Length of String = 52.8 cm
g = gravity = 9.81 m/s2
Substituting the values in equation (1),we get,
Notes:
The time period equation of simple harmonic motion is written as:Where,
m = mass of object
k = Springs constant
The restoring force of pendulum which accelerates the mass towards equilibrium is given as:
mg Sinφ = ks
s = rφ
In this case, r = L
Therefore, s = Lφ
Now,
mg Sinφ = k Lφ
k = mg Sinφ / Lφ
k = (mg/L) (Sinφ/φ)
For all small angles:
lim (φ--->0) = Sinφ/φ = 1
therefore. equation (2) implies:
In this question: φ = 33°
33° = 33 x π /180
33° = 0.575959 radians
Therefore,
Sin 0.575959 / 0.575959
0.9456
1/0.9456 = 1.057505
Sqrt (1.057505)
1.02835
Now the formula can be written as:
So, the accurate answer would be 1.50 s.
Ответ:
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