The maximum speed of the object under simple harmonic motion is 0.786 m/s.
The given parameters:
y = A sin(ωt + Ф)
where;
The maximum speed of the object is calculated as follows;
Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.
Learn more about simple harmonic motion here: brainly.com/question/17315536
Answer:
Explanation:
<h3>Work</h3>We know that the differential work made by the gas its defined as:
We can solve this by integration:
but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law
This give us
As n, R and T are constants
But the volume is:
Now, lets use the value from the problem.
The temperature its:
The ideal gas constant:
So:
We know that, for an ideal gas, the energy is:
where its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know
where is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:
Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t =
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w =
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = -
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }