#$&* course PHy 121 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** The vertical component of the tension in the string is equal to the weight m * g of the pendulum. At angle `theta from equilibrium we have T cos(`theta) = m * g so T = m * g / cos(`theta). The horizontal component of the tension is the restoring force. This component is T sin(`theta) = m * g * sin(`theta) / cos(`theta) = m * g * tan(`theta). For small angles tan(`theta) is very close to `theta, assuming `theta to be in radians. Thus the horizontal component is very close to m * g * `theta. The displacement of the pendulum is L * sin(`theta), where L is pendulum length. Since for small angles sin(`theta) is very close to `theta, we have for small displacements x = displacement = L * `theta. Thus for small displacements (which implies small angles) we have to very good approximation: displacement = x = L `theta so that `theta = x / L, and restoring force = m * g * `theta = m * g * x / L = ( m*g/L) * x. ** ALTERNATIVE EXPLANATION: This relationship can be easily visualized in terms of similar triangles. Sketches can also be found in the appropriate Class Notes. First: The angle of the pendulum with vertical is small. It follows that the vertical component of the tension force is very nearly the same as the tension force. The vertical component of the tension force is what prevents the mass at the end of the pendulum from falling. That mass doesn't move significantly in the y direction, so it is in vertical equilibrium. We conclude that the vertical component of the tension has magnitude m g, and that the tension therefore has magnitude m g. Now consider two triangles (which unless you have excellent powers of visualization you should sketch as you read the description): • The first triangle has as its hypotenuse the length of the pendulum, and its legs are horizontal and vertical. • The second triangle has as its hypotenuse the tension force acting on the pendulum mass, and its legs are horizontal and vertical. The tension force acts in the direction of the string, so the string makes the same angle with horizontal as the tension vector. We conclude that the triangles are similar. • The horizontal leg of the first triangle corresponds to the displacement of the pendulum in the horizontal direction. Call this leg x. • The horizontal leg of the second triangle corresponds to the component of the tension force in the horizontal direction. Call this leg T_horiz. • So the first triangle has horizontal leg x and hypotenuse L, where L is the length of the pendulum. • The second triangle has horizontal leg T_horiz and hypotenuse T = m g. Since the triangles are similar, we have the proportion • x / L = T_horiz / mg which we solve easily to find that • T_horiz = mg * (x / L) = (m g / L) * x. The horizontal component T_horiz of the tension is in the direction opposite the horizontal displacement x. The horizontal component of the tension is the force that restores the pendulum to equilibrium. Thus • restoring force = - (m g / L) * x, or • F_restoring = - k x, with k = m g / L &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q What does simple harmonic motion have to do with a linear restoring force of the form F = - k x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a mass has the force F=-ks, if its in equilibrium it will stay there. If not, the simple harmonic motion will have frequency omega=sqrt(k/m). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the essential relationship is F = - k x; doesn't matter if it's a pendulum, a mass on an ideal spring, or any other system where net force is a negative constant multiple of the displacement from equilibrium. For Principles of Physics and General College Physics students the following statement summarizes the relationship: • If a constant mass m is subjected to a net force F = - k x, then if that mass is in equilibrium at x = 0 it will remain there. • Otherwise it will undergo simple harmonic motion with angular frequency omega = sqrt( k / m ). The following characteristics of the motion of the oscillator are not expected at this stage, but they are worth noting and thinking about: • Its total energy will be 1/2 k A^2, where A is the amplitude of its motion (its maximum displacement from the x = 0 position). • As it passes its equilibrium position all its energy will all be kinetic. • At its maximum displacement from equilibrium its energy will all be potential. • Its potential energy at position x is 1/2 k x^2. The rest of its total energy is in the form of kinetic energy. University Physic students should also understand the reasons for this: x is regarded as a function of clock time t, and the expression x '' indicates the second derivative of x with respect to clock time t. F = m * a = m * x'', so F = - k x means that • m * x'' = - k x. The only real-valued functions whose second derivatives are constant negative multiples of the functions themselves are the sine and cosine functions. • We conclude that x = A sin(`omega t) + B cos(`omega t), where `omega = `sqrt(k/m). For appropriate C and `phi, easily found by trigonometric identities, A sin(`omega t) + B cos(`omega t) = C sin(`omega t + `phi). • Thus SHM can be modeled by a point moving around a reference circle of appropriate radius C, starting at position `phi when t = 0. ** STUDENT QUESTION My question is, how does this answer the question? Does omega determin F somehow? INSTRUCTOR RESPONSE The brief answer to the question is that when F_net = - k x, the object will undergo simple harmonic motion, and vice versa (i.e., if the object undergoes SHM, then F_net = - k x). The given solution goes into more detail about why this is so. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: For a simple harmonic oscillator of mass m subject to linear net restoring force F = - kx, what is the angular velocity `omega of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: omega= sqrt (k/m) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: omega= sqrt (k/m) INSTRUCTOR COMMENT: Good. Remember that this is a very frequently used relationship for SHM, appearing in one way on another in a wide variety of problems. STUDENT COMMENT: i got this one right, but i thought you wanted it in m/s INSTRUCTOR RESPONSE: omega is the angular velocity, which is in rad / sec. If you want the speed of the point on the reference circle, then you would multiply the angular velocity by the radius of the reference circle. So for example if the radius is in meters, the speed of the point would be in meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q If the angular position of the point on the reference circle is given at clock time t by `theta = `omega * t, then what is the x coordinate of that point at clock time t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = radius*cos(omega*t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since theta=omega t, if we know t we find that x = radius * cos(theta) or more specifically in terms of t x = radius*cos(omega*t) STUDENT QUESTION I understand that theta = omega*t was derived from the concept of velocity = distance /time. But I don't understand how they are finding the x coordinate. INSTRUCTOR RESPONSE A vector whose length is equal to the radius, which makes angle theta with the positive x axis, has x and y components radius * cos(theta) and radius * sin(theta), respectively. Thus x = radius * cos(omega * t) and y = radius * sin(omega * t). You should of course sketch all this out on paper. An unlabeled sketch depicting a typical vector and its x and y components is depicted in the figure below: STUDENT QUESTION To find the x coordinate on the circle, we rely on theta=omega t to help us out. If we know t, then we can use x= radius*cos(theta) which then becomes a=radius*cos(omega*t) I flipped through a chapter in the book as well and saw an example that related to positions as a function of time. It had x=Acos theta. Could this formula apply to this situation as well? Is it essentially the same formula or is it for something completely different? And what exactly does the A represent? INSTRUCTOR RESPONSE A represents the amplitude of motion, which is equal to the radius of the reference circle. It is the same formula. The radius of the reference circle is equal to the amplitude of the motion, and the angle is omega * t. So x = radius * cos(ometa * t) is exactly the same as x = A * cos(theta). Recall from Intro Prob Set 5 that the x component of a vector of magnitude A and angle theta is v_x = A cos(theta). So this is nothing new. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query introductory problem set 9, #'s 1-11 If we know the restoring force constant, how do we find the work required to displace the oscillator from its equilibrium position to distance x = A from that position? How could we use this work to determine the velocity of the object at its equilibrium position, provided we know its mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v = sqrt(k/m) * A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You can use the work 1/2 k A^2 and the fact that the force is conservative to conclude that the max PE of the system is 1/2 k A^2. This PE will have transformed completely into KE at the equilibrium point so that at the equilibrium point KE = .5 m v^2 = .5 k A^2. We easily solve for v, obtaining v = `sqrt(k/m) * A. ** STUDENT COMMENT: I'm a little confused by that 1/2 k A^2. INSTRUCTOR RESPONSE: That is the PE at x = A. To directly reason out the expression PE = .5 k A^2 we proceed as follows: PE = work done by system in moving from equilibirum * displacement = fAve * `ds. The force exerted on the system at position x = A is -k A. The force exerted at position x = 0 is 0. Force changes linearly with position. So the average force exerted on the system is ( 0 - kA) / 2 = -1/2 k A. The force exerted by the system is equal and opposite, so fAve = 1/2 k A. The displacement from x = 0 to x = A is `ds = A - 0 = A. We therefore have PE = fAve `ds = 1/2 k A * A = 1/2 k A^2. This is also the area beneath the F vs. x graph between x = 0 and x = A. That region is a triangle with base A and altitude k A so its area is 1/2 base * height = 1/2 k A^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!