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Phys 2214 hw1 Solutions

Course: PHYSICS 2214, Fall 2011
School: Cornell
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Word Count: 812

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2214 Phys Homework #1 Solutions September 1, 2011 1. z (t) = rei(t+) (a) z (t) = rei(t+) = r[cos(t + ) + i sin(t + )]. Then x(t) = Re[z (t)] = r cos(t + ) and y (t) = Im[z (t)] = r sin(t + ) . (b) Suppose = 0 + i1 . Then z (t) = rei(t+) = rei(0 +i1 t+) = re1 t ei(0 t+) and Re[z (t)] = re1 t cos(0 t + ). (c) i. ii. iii. dz = irei(t+) dt d d {Re[z (t)]} = dt [r cos(t dt Re dz dt + )] = r sin(t + ) =...

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2214 Phys Homework #1 Solutions September 1, 2011 1. z (t) = rei(t+) (a) z (t) = rei(t+) = r[cos(t + ) + i sin(t + )]. Then x(t) = Re[z (t)] = r cos(t + ) and y (t) = Im[z (t)] = r sin(t + ) . (b) Suppose = 0 + i1 . Then z (t) = rei(t+) = rei(0 +i1 t+) = re1 t ei(0 t+) and Re[z (t)] = re1 t cos(0 t + ). (c) i. ii. iii. dz = irei(t+) dt d d {Re[z (t)]} = dt [r cos(t dt Re dz dt + )] = r sin(t + ) = Re[irei(t+) ] = Re{ir[cos(t + ) + i sin(t + )]} = Re[ir cos(t + ) r sin(t + )] = r sin(t + ) By comparing (ii) with (iii), we see that mute. (d) Given the equation a d2 x dx + b + cx = 0 dt2 dt 1 d [ dt ] and Re[ ] com- If x(t) = Re[z (t)], then if we substitute this into the equation above and use the result from part (b) we get: Re{a d2 z dz + b + cz } = 0 . 2 dt dt In general we cant conclude that if Re[z (t)] = 0 then z (t) = 0, since it isnt necessary that Im[z (t)] = 0. However, in this case, substitute z (t) = rei(t+) into the previous equation to get Re[z (t)(a 2 + ib + c)] = 0 . Since z (t) contains the only time dependence, the only way for this equation to hold is if a 2 + ib + c = 0. But if we substitute z (t) = rei(t+) into the original equation written in terms of z (t) we get dz d2 z a 2 + b + cz = z (t)(a 2 + ib + c) , dt dt and since we know that a 2 + ib + c = 0 we have a d2 z dz + b + cz = 0 . 2 dt dt 2. ei = cos +i sin . Then the complex conjugate is ei = cos i sin . Adding these two equations gives ei + ei = 2 cos , or cos = ei + ei . 2 Subtracting the two above equations gives ei ei = 2i sin , or sin = ei ei . 2i 3. (a) See gure. (b) xeq = 0.15 mm, A = (0.2/mm 0.1/mm)/2 = 0.05 mm, T = 0.135 s, f = 1/T = 7.4 Hz, = 2f = 46.5 rad/s, and 0 = +/2, since the graph is a cosine shifted to the left by 900 or /2 radians. 2 0.005 0.010 0.015 0.020 2 4 6 4. (a) We can write the position dependence of the mass as x(t) = x0 cos(2/T )t = x0 cos t, where the angular frequency = 2/T = 62.83rad/s. The velocity is the time derivative of the position. v = x0 sin t. And the acceleration is the time derivative of the velocity. a = 2 x0 cos t. We know that F = ma = m 2 x0 cos t. The maximum magnitude of the force is when cos t = 1. Then Fmax = m 2 x0 = (0.1kg )(62.83s1 )2 (0.1m) N (b) = 39.5 The force F = kx. Therefore k = |Fmax /x0 | = 395 N/m . (c) The total mechanical energy is equal to the sum of potential energy and kinetic energy. The velocity, (and therefore the ki1 netic energy) is zero when t = 0 and Emech = U = 2 kx2 = 0 1 2 (395 N/m)(0.1m) = 1.98 J . 2 (d) See plots on the next page. 5. (a) The maximum force on the bed before it slips is Fmax = s FN = s mg If the bed does not slip, it too has acceleration a, and the force on the bed would be F = ma. If mamax > Fmax = s mg , the bed slips. That happens if amax > s g = (0.8)(10 m/s2 ) = 8 m/s2 (b) We suppose that the oor is oscillating horizontally with position dependence x = x0 cos t. The horizontal velocity is v = dx = dt x0 sin t and the acceleration is a = dv = 2 x0 cos t. We dt see from the comparison of the equations for velocity and acceleration that vmax = amax / . The angular frequency = 2/T = 2/(2s) = 3.14 rad/s, therefore vmax > (8 m/s2 )/3.14s1 = 2.5ms so vmax > 2.5ms. 3 4 6. (a) The natural frequency of the oscillator is 0 = 2f0 = 2 (100Hz) = 628 rad/s (b) Q = 0 / = f0 /f = (100/20) = 5 . (c) Since Q = 0 , = Q/0 = 5/(628/s) = 8 ms (d) If = 0 , then A(0 ) = Fmax /m 0 / If << 0 then A( Fmax /m 0 ) 4 0 + 2 / 2 Fmax /m 2 0 as long as is not too small. Then the ratio of the amplitude on resonance to far o resonance is A( = 0 ) = 0 = Q A( 0 ) (e) The maximum displacement decays exponentially with time. xmax = Aet/(2 ) . Then xmax /A et/(2 ) = 0.1 and t = (2 ) ln(0.1) = 2(0.008 s) ln(0.1) = 0.037 s. 7. The force is obtained from the potential energy using F = dU/dx, so for a given value of x, the force is the slope of the potential energy curve at that value of x. Looking at the two curves it is clear that the slope of the exact curve is greater thatn the slope of the approx. curve, so the restoring force due to the exact potential is greater than the restoring force due to the approx. potential. That means that the object will be moving faster for the exact potential energy and therefore the period will be shorter and therefore the frequency will be greater for larger amplitudes than for smaller amplitudes. 5 1.2 Exact Approx 1.0 0.8 0.6 0.4 0.2 0.5 0.5 6
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