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1 Solutions.4-Page 4 Problem 3 A mass of 3 kg is attached to the end of a spring that is stretched cm by a force of 5N. It is set in motion with initial position = and initial velocity v = m/s. Find the amplitude, period, and frequency of the resulting motion. The spring constants, k = 5N/. m = 75 N/m. There is no mention of damping in the problem statement, and no outside forces acting on the system. The equation that governs the motion of the mass is =. Dividing through by the mass + 5 = k ω, the circular frequency, is calculated as = 5 rad / s. m From eq. on pg.35, the solution of the above differential equation is of the following form: ( = Acos5t + Bsin 5t Two equations will be necessary to solve for the two unknowns, A and B. Differentiating the position function to get the velocity function will provide the second equation. ( = v( = 5B cos5t 5Asin 5t The initial conditions are now substituted () = = A v() = = 5B B = ( = sin 5t The amplitude is m. ω 5 Frequency = = =. 8 Hz π π Period = =. 6 s Frequency
2 Problem 8 Most grandfather clocks have pendulums with adjustable lengths. One such clock loses min per day when the length of its pendulum is 3 in. With what length pendulum will this clock keep perfect time? A relationship between the period and length of the pendulum must be developed for the two situations. According to eq.6 on pg.34, the circular frequency,ω, of a pendulum is given by ω = g / L = GM / R L. Therefore the period, p = π / ω = πr L / GM. Dividing this equation for period by another period with another length gives the necessary relationship for this problem: p = (This result is also given in Problem 5.) p L L The given information is as follows: L = 3 Assuming the pendulum eecutes n cycles per day (44 minutes in a day), its period when it s keeping perfect time is p = 44 / n minutes. Since the pendulum takes min longer to carry out the same number of cycles, the period when the pendulum loses min is p = 45 / n minutes. Solving for L L p 3(44) gives L = = p (45) L = 9.6 in
3 Problem Consider a floating cylindrical buoy with radius r, height h, and uniform density ρ. 5 (recall that the density of water is g/cm 3 ). The buoy is initially suspended at rest with its bottom at the top surface of the water and is released at time t =. Thereafter it is acted on by two forces: a downward gravitational force equal to its weight mg = ρπr hg and an upward force of buoyancy equal to the weight πr g of water displaced, where = ( is the depth of the bottom of the buoy beneath the surface at time t (Fig..4.9). Conclude that the buoy undergoes simple harmonic motion around its equilibrium position e = ρh with period p = π ρh / g. Compute p and the amplitude of the motion if ρ =. 5 g/cm 3, h = cm, and g = 98 cm/s. The F = ma is ρπr h = ρπr hg πr g, if the downward direction is taken to be positive. (Note: The upward buoyancy force should be ρ waterπr g in order to have the proper units of force. ρ is left out because ρ = for water.) The above equation simplifies to + ( g / ρh) = g. This is a nonhomogeneous equation. Recall from pg.8 that the solution of a nonhomogeneous equation is a particular solution plus the complimentary solution. = Acosω t + Bsinω t c o The particular solution is the equilibrium position. (Substitute e = ρh into the differential equation and you will see that it works.) Thus the general solution is: ( Acosω ot + Bsinω t + ρh = ω = g / ρh o Differentiating to get a second equation gives = Bω o cosω ot Aω o sinω ot. Substituting the initial conditions, ( ) = () =, and simplifying yields: ( = ρh( cosω o This is an equation for harmonic motion. With the given values for ρ, h, and g, the period is easily solved for using p = π ρh / g. p =.sec
4 Problem 4 Suppose that the mass in a mass-spring-dashpot system with m = 5, c =, and k = 6 is set in motion with ( ) = and ( ) = 4. (a) Find the position function ( and show that its graph looks as indicated in Fig..4.. (b) Find the pseudoperiod of the oscillations and the equations of the envelope curves that are dashed in the figure. (a) From the given information, the differential equation governing the motion of the mass is = + (.) = c From pg.37, p = =. and ω = k / m = 3. m From eq.7 on pg.37, the characteristic equation corresponding to this differential equation has roots / r, r = p ± ( p ω ) which depend on the sign of p ω. From pg.37, c 4km.4 4(3.)5 p ω = = < and thus the system is underdamped. 4m 4(5) For the underdamped case (pg.38), the characteristic equation has comple roots ± p ± i ω p =. 3i Therefore the solution of the governing differential equation is.t ( = e ( Acos3t + Bsin 3 Differentiating the position function will give the necessary second equation so that the initial conditions can be used to solve for A and B..t.t ( =.e ( Acos3t + Bsin 3 + e (3B cos3t 3Asin 3 Substituting the initial conditions gives the following system of equations: () = = A () = 4 =.() + (3B) B = 5 C = + 5 = 5 From eq., ( = 5e first quadrant.).t cos(3t α) where α = tan (5/ ) =.6435rad (α is in the
5 ( = 5e.t cos(3t.64) The graph is given below. The computer program Maple was used to graph the position equation. (b) The pseudoperiod is T = π / 3 From fig..4.8, the envelope curves are t = ±5e.
6 Problem 3 This problem deals with a highly simplified model of a car of weight 3 lb (mass m = slugs in fps units). Assume that the suspension system acts like a single spring and its shock absorbers like a single dashpot, so that its vertical vibrations satisfy Eq. (4) with appropriate values of the coefficients. (a) Find the stiffness coefficients k of the spring if the car undergoes free vibrations at 8 cycles per minute (cycles/min) when its shock absorbers are disconnected. (b) With the shock absorbers connected the car is set into vibration by driving it over a bump, and the resulting damped vibrations have a frequency of 78 cycles/min. After how long will the time-varying amplitude be % of its initial value? With m =, we have ω = k /. We are also given information about the frequency. Recall from Eq.4 on pg.35 that the relationship between frequency and circular frequency is ω = πν. ω = π (8 cycles/min)( min/6 sec) =8π / 3 Equating the known information on circular frequency yields: 8π / 3 = k / k = 78lb/ft (b) Now ω = π ( 78 cycles/min)( min/6 sec) = sec. 4km c Eq. on pg.38 states that ω = ω p =. Substituting the known m values and solving for the damping constant yields c = Therefore p = c / m =.865. Finally e =. gives t =. 47 sec
7 Problem 4 Problems 4-34 deal with a mass-spring-dashpot system having position function ( satisfying Eq.(4). We write o = () and v = () and recall that p = c /( m), ω = k / m, and ω = ω p. The system is critically damped, overdamped, or underdamped, as specified in each problem. (Critically damped) Show that in this case that ( + v t + p t e ( ) pt = pt From eq.9 on pg.38, = e ( c + c ). Differentiating yields pt ( c + c + c = pe constants. v = c = p( ) + c c = v + p e ( t. Now the differential equations can be used to solve for the Substituting the for the constants gives ( ( + v t + p e =
8 Problem 34 Problems 4-34 deal with a mass-spring-dashpot system having position function ( satisfying Eq.(4). We write o = () and v = () and recall that p = c /( m), ω = k / m, and ω = ω p. The system is critically damped, overdamped, or underdamped, as specified in each problem. (Underdamped) A body weighing lb (mass m = 3. 5 slugs in fps units) is oscillating attached to a spring and a dashpot. Its first two maimum displacements of 6.73 in. and.46 in. are observed to occur at times.34 s and.7 s, respectively. Compute the damping constant (in pound-seconds per foo and spring constant (in pounds per foo. Note that the problem does not state whether the maimum displacements occur approimately a pseudoperiod apart or approimately a half pseudoperiod apart. If they are approimately a full pseudoperiod apart then the displacements are positive. However, if they are a half pseudoperiod apart then one of the displacements would be negative. For an underdamped system ( = Ce cos( ωt α), then pt ( = pce cos( ωt α) + Cωe sin( ωt α). The derivative is set to zero because maimums (or minimums) occur when the derivative is zero. Simplifying yields tan( ω t α) = p / ω. Thus at a maima, the tan function has a constant value. The tangent of an angle is the same for another angle only if the angles are π apart. Thus ω t ωt = π (see the result of Problem 3). This equation can be used along with the given times when the maimums occur to find that ω = The result of Problem 33 is p =.84 πp ln =. With 73 ω = 6. and =. 46, this leads to Eq.6 on pg.37 gives c = mp = ( / 3)(.84) =. 5lb-sec/ft And Eq. yields k = (4m ω + c ) / 4m = lb/ft
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