Respuesta :
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 [tex]m/s^2[/tex]
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:
[tex]x = Asin \omega t[/tex]
Differentiating the above expression with respect to x ; we have :
[tex]\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)[/tex]
[tex]v = A \omega cos \omega t[/tex]
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:
[tex]v_{max} = A \omega[/tex]
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 [tex]2 \pi f[/tex] for [tex]\omega[/tex] in the above expression;
[tex]v_{max} = A(2 \pi f)[/tex]
[tex]v_{max} = 3.40 cm (2 \pi *4.80)[/tex]
[tex]v_{max} = 102.54 \ cm/s[/tex]
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
[tex]v = A \omega cos \omega t[/tex]
By differentiation with respect to t;
[tex]\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)[/tex]
[tex]a =- A \omega^2 sin \omega t[/tex]
The maximum acceleration of the sphere is;
[tex]a_{max} =A \omega^2[/tex]
where;
[tex]w = 2 \pi f[/tex]
[tex]a_{max} = A(2 \pi f)^2[/tex]
where A= 3.40 cm = 0.034 m
[tex]a_{max} = 0.034*(2 \pi *4.80)^2[/tex]
[tex]a_{max} = 30.93 \ m/s^2[/tex]
The maximum magnitude of the acceleration of the sphere is = 30.93 [tex]m/s^2[/tex]
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 [tex]x = \pm A[/tex]