The amplitude of a lightly damped harmonic oscillator decreases from 60.0 cm to 40.0 cm in 10.0 s. What will be the amplitude of
the harmonic oscillator after another 10.0 s passes?
1 answer:
You might be interested in
The distance (ft) traveled by the particle at time t (s) is
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
The first law of thermodynamics says that the variation of internal energy of a system is given by:
where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is
where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is
Answer:
Explanation:
We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.
The x-component of the length of the minute hand is:
(1)
- d is the length of the minute hand (d=D/2)
- D is the diameter of the clock
- t is the time (min)
Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:
(2)
Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:
Now, let's put this value on (2)
Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:
I hope it helps you!