#$&* course Phy 121 034. `query 34*********************************************
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Given Solution: `a** The vertical component of the tension in the string is equal to the weight m * g of the pendulum. At angle `theta from equilibrium we have T cos(`theta) = m * g so T = m * g / cos(`theta). The horizontal component of the tension is the restoring force. This component is T sin(`theta) = m * g * sin(`theta) / cos(`theta) = m * g * tan(`theta). For small angles tan(`theta) is very close to `theta, assuming `theta to be in radians. Thus the horizontal component is very close to m * g * `theta. The displacement of the pendulum is L * sin(`theta), where L is pendulum length. Since for small angles sin(`theta) is very close to `theta, we have for small displacements x = displacement = L * `theta. Thus for small displacements (which implies small angles) we have to very good approximation: displacement = x = L `theta so that `theta = x / L, and restoring force = m * g * `theta = m * g * x / L = ( m*g/L) * x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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 constant mass isn’t in equilibrium it will undergo harmonic motion with omega=sqrt(k/m) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the essential relationship is F = - k x; doesn't matter if it's a pendulum, a mass on an ideal spring, or any other system where net force is a negative constant multiple of the displacement from equilibrium. For Principles of Physics and General College Physics students the following statement summarizes the relationship: • If a constant mass m is subjected to a net force F = - k x, then if that mass is in equilibrium at x = 0 it will remain there. • Otherwise it will undergo simple harmonic motion with angular frequency omega = sqrt( k / m ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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 #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT RESPONSE: omega= sqrt (k/m) INSTRUCTOR COMMENT: Good. Remember that this is a very frequently used relationship for SHM, appearing in one way on another in a wide variety of problems. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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=r*cos(omega*t) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since theta=omega t, if we know t we find that x = radius * cos(theta) or more specifically in terms of t x = radius*cos(omega*t) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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: KE=1/2mv^2=1/2kA^2 V=sqrt(k/m)*A confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You can use the work 1/2 k A^2 and the fact that the force is conservative to conclude that the max PE of the system is 1/2 k A^2. This PE will have transformed completely into KE at the equilibrium point so that at the equilibrium point KE = .5 m v^2 = .5 k A^2. We easily solve for v, obtaining v = `sqrt(k/m) * A. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. Some of the relationships I discovered surprised me. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "