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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 distanc

geniusboy [140]

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 $m/s^2$

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:

$x = Asin \omega t$

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 $x = \pm A$

6 0

3 years ago

If you were to yell at a canyon wall, you would hear yourself a short time later. This is because sound can be

mojhsa [17]

The answer is answer b refracted

5 0

3 years ago

Read 2 more answers

Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert

chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h

5 0

3 years ago

Read 2 more answers

Earth has four motions in its movement through space, rotation, revolution, processional, and solar motion, which two are of any

Anna007 [38]

Answer:

rotation and revolution

Explanation:

out of the four motions the earth is subject to which are: rotation about its axis, revolution around the Sun, processional motion (a slow conical movement ) of the axis, and the solar motion (this refers to the

movement of the whole solar system with space), only two are of any

importance to meteorology as this two causes changes in weather and seasons. The first motion is rotation. Earth rotates on its axis

once every 24 hours. One-half of the Earth’s surface is

therefore facing the Sun at all times. The second motion of Earth is its revolution around the Sun. The revolution around the Sun and the earth tilt on its axis are responsible for changes in seasons. The Earth

makes one complete revolution around the Sun in

approximately 365 1/4 days.

6 0

4 years ago

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago