A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
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Answer:
a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b) The maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) The maximum magnitude of the acceleration of the sphere is = 30.93
The maximum acceleration occurs at maximum excursion from equilibrium.
Explanation:
Given that :
Frequency (f) = 4.80 Hz
Amplitude (A) = 3.40 cm
a)
The total distance traveled by the sphere during one cycle of simple harmonic motion is:
d = 4A (where A is the Amplitude)
d = 4(3.40 cm)
d = 13.60 cm
Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b)
As we all know that:
Differentiating the above expression with respect to x ; we have :
Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;
Then:
We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium
substituting for
in the above expression;
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
By differentiation with respect to t;
The maximum acceleration of the sphere is;
where;
where A= 3.40 cm = 0.034 m
The maximum magnitude of the acceleration of the sphere is = 30.93
The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of