Physics, 05.05.2020 20:05 whitakers87

A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distance (in cm) does the sphere move during one cycle of its motion? cm (b) What is its maximum speed (in cm/s)? cm/s Where does this maximum speed occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these (c) What is the maximum magnitude of acceleration (in m/s2) of the sphere? m/s2 Where in the motion does the maximum acceleration occur? as the sphere passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these

Answers: 2

Show answers

Answers

Answer from: vnzgirl

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

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

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

As we all know that:

$x = Asin \omega t$

Differentiating the above expression with respect to x ; we have :

$\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)$

$v = A \omega cos \omega t$

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:

$v_{max} = A \omega$

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 $2 \pi f$ for $\omega$ in the above expression;

$v_{max} = A(2 \pi f)$

$v_{max} = 3.40 cm (2 \pi *4.80)$

$v_{max} = 102.54 \ cm/s$

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

$v = A \omega cos \omega t$

By differentiation with respect to t;

$\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)$

$a =- A \omega^2 sin \omega t$

The maximum acceleration of the sphere is;

$a_{max} =A \omega^2$

where;

$w = 2 \pi f$

$a_{max} = A(2 \pi f)^2$

where A= 3.40 cm = 0.034 m

$a_{max} = 0.034*(2 \pi *4.80)^2$

$a_{max} = 30.93 \ m/s^2$

The maximum magnitude of the acceleration of the sphere is = 30.93 m/s^2

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 $x = \pm A$

Answer from: Quest

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2 y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x | y

-2 | 2

0 | 8

2 | 14