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 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
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:
![x = Asin \omega t](https://tex.z-dn.net/?f=x%20%3D%20Asin%20%5Comega%20t)
Differentiating the above expression with respect to x ; we have :
![\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28Asin%20%5Comega%20t%29)
![v = A \omega cos \omega t](https://tex.z-dn.net/?f=v%20%3D%20A%20%5Comega%20cos%20%5Comega%20t)
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:
![v_{max} = A \omega](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20A%20%5Comega)
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;
![v_{max} = A(2 \pi f)](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20A%282%20%5Cpi%20f%29)
![v_{max} = 3.40 cm (2 \pi *4.80)](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%203.40%20cm%20%282%20%5Cpi%20%2A4.80%29)
![v_{max} = 102.54 \ cm/s](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20102.54%20%5C%20cm%2Fs)
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
![v = A \omega cos \omega t](https://tex.z-dn.net/?f=v%20%3D%20A%20%5Comega%20cos%20%5Comega%20t)
By differentiation with respect to t;
![\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%28v%29%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%28A%20%5Comega%20cos%20%5Comega%20t%29)
![a =- A \omega^2 sin \omega t](https://tex.z-dn.net/?f=a%20%3D-%20A%20%5Comega%5E2%20sin%20%5Comega%20t)
The maximum acceleration of the sphere is;
![a_{max} =A \omega^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3DA%20%5Comega%5E2)
where;
![w = 2 \pi f](https://tex.z-dn.net/?f=w%20%3D%202%20%5Cpi%20f)
![a_{max} = A(2 \pi f)^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20A%282%20%5Cpi%20f%29%5E2)
where A= 3.40 cm = 0.034 m
![a_{max} = 0.034*(2 \pi *4.80)^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%200.034%2A%282%20%5Cpi%20%2A4.80%29%5E2)
![a_{max} = 30.93 \ m/s^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%2030.93%20%5C%20m%2Fs%5E2)
The maximum magnitude of the acceleration of the sphere is = 30.93 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
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 ![x = \pm A](https://tex.z-dn.net/?f=x%20%20%3D%20%5Cpm%20A)