HW5 Physics 311 Mechanics

Transcription

1 HW5 Physics Mechanics Fall 05 Physics department University of Wisconsin, Madison Instructor: Professor Stefan Westerhoff By Nasser M. Abbasi June, 06

2 Contents 0. Problem Problem Part ( Part ( Problem Problem Part ( Part ( Part ( Problem Part( Part(

3 Fall 05 Homework 5 (0/9/5, due 0//5 0. Problem. (5 points A spring of spring constant k supports a box of mass M, which contains a block of mass m. If the system is pulled downward a distance d from the equilibrium position and then released, it starts to oscillate. For what value of d does the block just begin to leave the bottom of the box at the top of the vertical oscillations? SOLUTION:. (5 points ( Show that the Fourier series of a periodic square wave is f(t 4 [sin(t+ inital conditions π sin(t+ ] y(0 d, k 5 sin(5t+... ẏ(0 0, where M f(t + m for 0 < t < π, π < t < π,... f(t for π < t y< π, π < t < 4π,.... ( Use the result from above to find the steady-state motion of a damped harmonic The block oscillator of mass that m will is driven leave the by afloor periodic of thesquare-wave box when the force vertical of amplitude acceleration F 0. isinlarge particular, enough to match findtherelativeamplitudesofthefirstthreeterms, gravity acceleration g. The equation motion A, of A the, anda overall 5, oftheresponsefunction system is given by x(t in the case that the third harmonic of the driving frequency coincides with the y + frequency 0 of the undamped oscillator. 0y 0 ( Assume a quality factor of Q 00. Where 0 is the undamped natural frequency k. (5 points 0 M + m If the solar system were imbedded in a uniform dust cloud of density ρ, what would be the The solution force on to a( planet is a distance r from the center of the Sun? y A cos 0 t + B sin 0 t ( Initial conditions are used to find A, B. Since at t 0, y (0 d, then from ( we find Taking...continued derivative of on( next gives page... A d y A 0 sin 0 t + B 0 cos 0 t ( At t 0, y (0 0, this gives B 0. Therefore the full solution ( becomes y d cos 0 t The acceleration is now found as y 0 d sin 0 t y 0d cos 0 t

4 The period is T p π 0. After one T p from release the box will be the top. Therefore, the acceleration at that moment is y (T p 0d cos 0 T p 0d cos π 0d The condition for m to just leave the floor of the box is when the above acceleration is the same as g. 0d g d g 0 Therefore d g (M + m k 4

5 . (5 points A spring of spring constant k supports a box of mass M, which contains a block of mass m. If the system is pulled downward a distance d from the equilibrium position and then released, it starts to oscillate. For what value of d does the block just begin to leave the bottom of the box at the top of the vertical oscillations? 0. Problem. (5 points ( Show that the Fourier series of a periodic square wave is f(t 4 [sin(t+ π sin(t+ ] 5 sin(5t+..., where f(t + for 0 < t < π, π < t < π,... f(t for π < t < π, π < t < 4π,.... ( Use the result from above to find the steady-state motion of a damped harmonic oscillator that is driven by a periodic square-wave force of amplitude F 0. In particular, findtherelativeamplitudesofthefirstthreeterms, A, A, anda 5, oftheresponsefunction x(t in the case that the third harmonic of the driving frequency coincides with the frequency 0 of the undamped oscillator. Assume a quality factor of Q 00. SOLUTION:. (5 points If the solar system were imbedded in a uniform dust cloud of density ρ, what would be the 0.. force Part on ( a planet a distance r from the center of the Sun? The function f (t is an odd function, therefore we only need to evaluate b n terms. To more clearly see the period, the definition of f (t is written as { + 0 < t < π f (t, π...continued on next page... < t < π, Therefore the period is T p π 5

6 Finding b n b n T p π Tp 0 f (t sin (nt dt ( π (+ sin (nt dt + 0 ( π sin (nt dt π 0 ( [ ] π cos (nt π n 0 π ( n ( π π [ π [cos (nt] 0 + n [ cos ( n π cos (0 π π sin (nt dt cos (nt n ( sin (nt dt ] π π π [cos (nt] π ] + [ cos ( n π ( [cos ( ] + [cos ( cos (] ( cos ( + + cos ( cos ( {}}{ cos ( + cos ( + ( cos ( And since n is an integer, then cos ( ( n and the above reduces to ( cos n π ] Therefore Hence b n f (t b n ( ( n { 4 n,, 5, 0 otherwise n,,5, n,,5, b n sin (nt 4 sin (nt Writing down few terms to see the sequence f (t {sin (t + sin (t + sin (5t + sin (7t + } π 6

7 0.. Part ( When the system is driven by the above periodic square wave of amplitude F 0, the steady state response is the sum to the response of each harmonic in the Fourier series expansion of the forcing function. Since the steady state response of a second order system to F n sin (nt is given by y n (t F n /m ( 0 (n sin (nt + δ n + 4λ n (n Where the phase δ n is defined as δ n tan λ (n 0 (n Then the steady state response to f (t n,,5, F0 4 sin (nt is given by y ss (t n,,5, 4F 0 πm 4 n,,5, F 0 /m ( 0 (n sin (nt + δ n ( + 4λ (n n sin (nt + δ n ( 0 (n + 4λ (n Looking at the first three responses gives y ss (t 4F 0 sin (t + δ + sin (t + δ πm (0 + 4λ ( 0 ( + + 4λ ( ( We are told that 0 or 0 and in addition, using using Q 0 λ 00 0 λ λ 0 00 Using this λ and given value of then the phase δ n becomes δ n tan λ (n 0 (n tan ( 0 00 ( n 0 0 ( n 0 tan n 00n sin (5t + δ 5 ( 0 ( λ (5 we find 7

8 Using the above phase in ( gives y ss (t 4F 0 πm 4 π 4 π F 0 m F 0 m sin ( 0 t + π sin ( 0 t + tan 800 ( 0 ( + 0 ( ( 0 ( 00 0 ( + 0 ( ( 00 0 sin ( 0 t tan sin ( 0 t + π + sin ( 5 0 t + tan { sin ( 0. 0 t tan The relative amplitudes of A, A, A 5 are given by 0 +. sin ( 0 t + π {.5,., 0.49} sin ( 5 0 t + ( 0 ( sin ( t + tan 0 We see that the third harmonic (n has the largest amplitude, since this is where 0. In normalized size, dividing all amplitudes by the smallest amplitude gives {A, A, A 5 } normalized {0, 96, } 0 + The third harmonic n has π phase since tan ( π 8

9 ( Use the result from above to find the steady-state motion of a damped harmonic oscillator that is driven by a periodic square-wave force of amplitude F 0. In particular, findtherelativeamplitudesofthefirstthreeterms, A, A, anda 5, oftheresponsefunction x(t in the case that the third harmonic of the driving frequency coincides with the frequency 0 of the undamped oscillator. Assume a quality factor of Q Problem. (5 points If the solar system were imbedded in a uniform dust cloud of density ρ, what would be the force on a planet a distance r from the center of the Sun? SOLUTION:...continued on next page... dust cloud ρ solar system M s, R s sun r m planet The total force on the planet m is due to the mass inside the region centered at the center of the sun. The mass outside can be ignored since its effect cancels out. Let the radius of the sun be R sun, then the total mass that pulls the planet toward the center of the solar system is given by The force on the planet is therefore M total M sun + 4 π ( r R sun ρ F GM totalm ˆr r G ( M sun + 4π (r Rsun ρ m ˆr r Where ˆr is a unit vector pointing from the sun towards the planet m and G is the gravitational constant and ρ is the cloud density. 9

10 0.4 Problem 4 4. (0 points ( What is the speed (in km/s for a satellite in a low-lying orbit close to Earth? Assume that the radius of the satellite s orbit is roughly equal to the Earth s radius. ( Show that the radius for a circular orbit of a synchronous (4-h Earth satellite is about 6.6 Earth radii. ( The distance to the Moon is about 60. Earth radii. From this, calculate the length of the sidereal month (the period of the Moon s orbital revolution. SOLUTION: 5. (5 points ( A particle is subject to an attractive force f(r, where r is the distance between the 0.4. particle Part ( and the center of the force. Find f(r if all circular orbits are to have identical The force areal onvelocities. the satellite is mr e where r e is taken as the earth radius since this is low-lying orbit. Therefore ( The orbit of a particle moving in a central field is a circle passing through the origin, r r 0 cos(θ. Show that the forcegm law e mis inverse-fifth power. mr e But v r e where v is the satellite speed we want to find. Hence v r e r e and the above becomes GM e r e r e v r e GMe v r e ( ( meter/sec 7.9 km/sec 0.4. Part ( Let the radius of the satellite orbit be r. Using GM e m mr r ( GMe r 0

11 where π T p where T p is the period of the satellite. But for synchronous satellite, this period is 4 hrs. Hence the above becomes π T p r GM e ( ( ( ( π 4(60( meter But radius of earth is r e meters. Hence 0.4. Part ( r r e π T p r 60.r e moon T p is sidereal period earth (r e From GM e m mr r GM e r ( GM e π r T p We solve for T p, hence π GMe T p r T p π GM e r sec π ( ( ((60.(

12 Therefore, in days, the above becomes T p (4 (60 ( days

13 that the radius of the satellite s orbit is roughly equal to the Earth s radius. ( Show that the radius for a circular orbit of a synchronous (4-h Earth satellite is about 6.6 Earth radii. ( The distance to the Moon is about 60. Earth radii. From this, calculate the length of the sidereal month (the period of the Moon s orbital revolution. 0.5 Problem 5 5. (5 points ( A particle is subject to an attractive force f(r, where r is the distance between the particle and the center of the force. Find f(r if all circular orbits are to have identical areal velocities. ( The orbit of a particle moving in a central field is a circle passing through the origin, r r 0 cos(θ. Show that the force law is inverse-fifth power. SOLUTION: 0.5. Part( θ r ds A ( θ r m θ r m ds ( θ A r From the above diagram, where we have two particles of same mass m in two circular orbits. The area of each sector is given by The time rate of each sector area is Similarly A θ r da dt da dt θ r ( θ r ( Since we have a central force, then this force attracts each mass with a force given by f mr θ. Therefore f r mr θ, Similarly f r mr θ. Substituting for θ from these expressions back into ( and ( gives da f r (B dt mr Similarly da dt f r mr (B

14 We are told the areal speeds are the same, therefore equating the above gives da dt f mr r da dt f r mr f r 4 mr 4 f r mr 4 f r f r Hence f r r f r r This says that, since we using the same mass, that the force f (r on a mass is inversely proportional to the cube of the mass distance from the center. To see this more clearly, let r then f r r f r So if we move the mass from r to say times as far to r, then the force on the same mass becomes smaller than it was Part( The orbit first is plotted as follows Clear[r0, r] r0 ; r[angle_] : r0 Cos[angle] xydata Table[{r[a] Cos[a], r[a] Sin[a]}, {a, 0, Pi,.}]; ListLinePlot[xyData, GridLines -> Automatic, GridLinesStyle -> LightGray, AxesOrigin -> {0, 0}, AxesLabel -> {x, y}, BaseStyle -> 4, PlotTheme -> "Classic", AspectRatio -> Automatic] Which produces the following plot y Out[7] x

15 Using 8. in textbook, page 9 d dθ ( + r r µr F (r ( l Where µ is the reduces mass, l is the angular momentum and F (r is the force we are solving for. Since r r 0 cos θ then ( d d ( d d ( d dθ r dθ dθ r dθ dθ r 0 cos θ d ( ( ( sin θ dθ r 0 cos θ d ( sin θ dθ r 0 cos θ ( cos θ r 0 cos θ + sin θ r 0 cos θ ( r 0 cos θ + sin θ ( r 0 cos θ But from r r 0 cos θ we see that cos θ r r 0 Therefore ( becomes d dθ ( r ( and sin θ cos r, θ r 0 hence ( becomes ( + r r 0 r 0 ( r + r r 0 r r 0 r + r r0 r r 0 ( ( r + r 0 r r r + r 0 r r r 0 r r ( r r 0 ( r r 0 r 0 r0 r + r µr F (r r l r0 r + r µr r l F (r 5

16 Solving for F (r F (r l r0 µr ( 5 l r0 µ r 5 The above shows that the force is an inverse fifth power. 6