course Phy 201 S{瀩CeHẔ{assignment #034
......!!!!!!!!...................................
01:32:23 Query Class Notes #33 Why do we say that a pendulum obeys a linear restoring force law F = - k x for x small compared to pendulum length?
......!!!!!!!!...................................
RESPONSE --> When x is small, the angle of the pendulum with vertical is small and almost all of its displacement back toward equilibrium will be in the x direction.
.................................................
......!!!!!!!!...................................
01:40:00 ** 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. **
......!!!!!!!!...................................
RESPONSE --> The vertical component of the tension in the string is equal to m*g. At an angle of theta from equilibrium we have T(cos)(theta)=m*g T=m*g/cos(theta) For the horizontal component= T(sin)(theta)=m*g*sin theta/cos theta= m* g * tan(theta) For small angles x= displacement= L * theta theta=x/L restoring force= m*g * theta= m* g*x/L =(m*g/L)*x
.................................................
......!!!!!!!!...................................
01:48:50 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
......!!!!!!!!...................................
RESPONSE --> restoring force for pendulum= (m*g*x/L) F is the restoring force Tx and k is the quantity mg/L
.................................................
......!!!!!!!!...................................
01:50:59 ** 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. F = m * a = m * x'', so F = - k x means that m * x'' = - k x. The only 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), showing that SHM can be modeled by a point moving around a reference circle of appropriate radius C, starting at position `phi when t = 0. **
......!!!!!!!!...................................
RESPONSE --> F= m*a=m*x so that F=-kx means that m*x is equal to -kx
.................................................
......!!!!!!!!...................................
01:57:34 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?
......!!!!!!!!...................................
RESPONSE --> omega= sqrt of k/m for a pendulum, omega= sqrt of g/L
.................................................
......!!!!!!!!...................................
01:58:01 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.
......!!!!!!!!...................................
RESPONSE --> omega= sqrt of k/m
.................................................
......!!!!!!!!...................................
02:03:14 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?
......!!!!!!!!...................................
RESPONSE --> omega= sqrt of k/m theta= sqrt of k/m * t x= A cos (sqrt of k/m * t) x= A cos theta
.................................................
......!!!!!!!!...................................
02:04:22 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)
......!!!!!!!!...................................
RESPONSE --> x= radius*cos(theta) x= radius*cos(omega*t)
.................................................
nxn҈ԩijÅߺڑ assignment #034 c]x Physics I 04-17-2008
......!!!!!!!!...................................
13:08:55 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?
......!!!!!!!!...................................
RESPONSE --> Force= K * displacement from equilibrium Force /2= Fave Work= Fave*`ds KE= .5mv^2 .5mv^2=Fave*`ds Solve for v
.................................................
......!!!!!!!!...................................
13:14:34 Query Add comments on any surprises or insights you experienced
......!!!!!!!!...................................
RESPONSE --> v=`sqrt (k/m) *A Work= .5kA^2=max PE of the system KE=.5mv^2=.5kA^2
.................................................
......!!!!!!!!...................................
13:15:17 as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Work= max PE of the system I thought work would be equal to the kinetic energy of the system
.................................................