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Reading Assignment 3

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Mario Bayubaskoro

on 26 January 2015

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Transcript of Reading Assignment 3

Reading Assignment 3
Solutions Explained
x(t) = A cos(ωt + Φ)

At the starting condition of an oscillator, t = 0. Subsituting that yields:

x(0) = A cos(ω(0) + Φ) = A cos(Φ)

Since A is constant, initial position of the oscillator is determined by Φ.
1) The starting conditions of an oscillator are characterized by
a) the phase constant
9) Which statement is INCORRECT about energy in a simple harmonic oscillator, such as a mass oscillating on a spring like in Fig. 13-20?
c) The potential energy of the system is half the total energy at x = A/2
Don't forget that PE = 1/2 kx² so at x = A/2, the PE is actually 1/4 of the total energy and not 1/2.

Now let's see why the rest are correct:
a) The total energy, E, at position x=A and x= A/2 is the same
- we know that energy can neither be created nor destroyed so
total energy is the same at all times.
b) The potential energy of the system is equal to the total energy minus the kinetic energy, U = E - 1/2 mv²
-
E = K + U
U = E - K = E - 1/2 mv²
d) At position x = 0 the total energy is equal to the kinetic energy, E = K = 1/2 mv²
-
E = 1/2 kx² + 1/2 mv² = 1/2 k(0)² + 1/2mv² = 1/2mv²
e) At position x = A the total energy is equal to the potential energy, E = U = 1/2 kx2.
-
E = 1/2 kx² + 1/2 mv² = 1/2 k(A)² + 1/2m(0)² = 1/2k(A)²
2) An object undergoing simple harmonic motion has its maximum speed when
a) the object passes through x = 0
We can think of this in terms of energy conservation. At x = 0, potential energy is 0 and therefore all of the energy is kinetic energy.
E = PE + KE = 0 + 1/2 mv² = 1/2 mv²
3) An object undergoing simple harmonic motion has its maximum positive acceleration when
d) the object is at its maximum negative displacement (x = -A)
This can be explained using the concept of Hooke's law:

F = -kx

At maximum negative displacement, force is maximum and positive therefore acceleration is also maximum and positive.
4) An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is
a) unchanged
Max speed = ωA
Since ω = 2π/T, we can rewrite this as
Max speed = 2π/T x A

If we were to double the amplitude and period:
Max speed = 2π/(2T) x (2A)
The twos will cancel and therefore max speed will remain the same
5) Looking at the figure for Checkpoint C13-5, what is the phase constant Φ, of the ball in part (b)?
d) π
In the case of circular motion, the phase constant Φ represents the starting angle

As can be seen quite easily from the figure, the angle Φ with respect to the x axis is π
6) Imagine you compress a spring by small distance d. For this, you have to do some work W on the spring. How much work do you have to do if you compress the spring by a distance 3 d (three times as much)?
b) 9 W
Possibly a common error would be picking d) 3 W, thinking that W = Fd and if we were to increase d three times as much, then the work would also be increased three times as much. However, remember that the
force of the spring is not constant.

Through calculus, we are able to obtain the formula:
W = 1/2(kx²) assuming initial position = 0

Let
w = 1/2 (kd²)
and
W = 1/2 [k(3d)]²

Dividing these two together:
W/w = [1/2 (k(3d)²)] / [1/2 (kd)] = 3² = 9
Therefore W = 9 w
7) An object undergoing simple harmonic motion (see Fig 13-16) has an acceleration directed to the right (positive x-direction)
c) when it is at a negative x-position (to the left of zero)

Since there is an acceleration directed to the right (positive x-direction), there must exist a force, more specifically a
restoring force
that attempts to restore the object back to equilibrium to the right. Therefore, the object must have been located to the left of the equilibrium which is a negative x position.
8) A block (0.2 kg) attached to a horizontal spring is pulled 20 cm and released. Which statement is NOT true about the energy in the system just BEFORE the block is released?
d) At the displacement maxima, all the energy is stored as potential energy due to gravity, so Etotal = U = mgh
Since the mass is not moving in the vertical direction, gravity does not affect it movement. Instead, at the displacement maxima, all the energy is stored as electric potential energy.

Now let us see why the rest are correct:
a) At the displacement maxima, all the energy is stored as potential energy of the spring, so Etotal = U = 1/2kA²
- This is true because at displacement maxima,
velocity = 0 and so E = KE + U = 0 + U = 1/2kA²

b) Etotal ALWAYS equals K + U, and at the maxima K = 0.
- we know that Total Energy = Kinetic Energy + Potential Energy at all times. At
maxima
, the
velocity = 0
so kinetic energy is also 0

c) If given the spring constant k = 10 N/m, it would be possible to solve for the maximum velocity of this block.
-
Total energy = total energy
(in other words, total energy is the same at all times)
velocity is max when x = 0. Therefore, we may have this equation:
1/2 mv² = 1/2 kx²
=> left side represents situation of max velocity and right side represents inital stage (spring fully stretched with mass at rest). You are given everything except the v, so you can easily solve for it.
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