1, wave, 19, harmonic, 40, 48

Ch 15.1 #1

A ball dropped from a height of 4.00 m makes a perfectly elastic collision with the ground. Assuming no mechanical energy is lost due to air resistance, (a) show that the ensuing motion is periodic and (b) determine the period of the motion. (c) Is the motion simple harmonic? Explain.

a) Since the collision is elastic, no energy is transferred out of the system during the rebound. We always ignore air friction unless told otherwise…thus the ball will continue to repeat its bounce to 4 meters every bounce.	b) d = ½at^2 t = √(2d/a) tdown = 0.9035 sec T = 1.807 sec		c) the force is mg not kx…so the force isn’t dependant upon position, thus not sinusoidal, thus not simple harmonic.
This top diagram is the displacement of a spring. While the spring is going AWAY from equilibrium, we have a negative acceleration. A further explanation of (c). Let’s look at the restoring force vector (hence acceleration, F = ma) on the position versus time diagram. This bottom diagram is a displacement of a ball. As the ball goes down…it’s accelerating and as it’s going up…it’s slowing down.

Ch 15.2 wave

A particle with an amplitude of 6.00 cm, and a frequency 4.00 Hz is moving along the x axis in simple harmonic motion. The particle starts from its equilibrium position, the origin, at t = 0 and moves to the right.

(a) Show that the position of the particle is given by x = (6.00 cm) sin(8πt)

(b) Determine the maximum speed and the earliest time (t > 0) at which the particle has this speed, (c) the maximum acceleration and the earliest time (t > 0) at which the particle has this acceleration, and (d) the total distance traveled between t = 0 and t = 1 second.

(a) Is ω correct? ω = 2πf ω = 2π(4) ω = 8π Yes Is the particle at origin when t = 0; x(0)=6 sin(8π(0)) x(0) = 0 Yes	@ t = 0; v is + v = d x /dt v(0) = d(6sin(ωt))/dt v(0) = 6ω cos(ωt) v(0) = +6ω Yes thus x(t) = 2 sin(3π t)		(b) vmax = A ω vmax = 6(8π) vmax = 151 cm/s		peaks occur when derivative = 0 vmax occurs at t = 0 and every ½T beyond.
			dv/dt = -A ω^2 sin(ωt) 0 = -A ω^2 sin(ωt) t = 0
			If T = 1/f = 1/4 sec, then next peak is @ ½ T = 1/8 seconds
(c) a = dv/dt = -6 ω^2 sin(ωt) amax = +6 ω^2 amax = +6 (8π)^2 amax = 3790 cm/s^2 da/dt = -6 ω^3 cos(ωt) peaks occur when 0 = -6 ω^3 cos(ωt) derivative = 0 π/2 = ωt t = 1/16 sec (this is 1^st negative accel peak) t = 1/16 + 1/8 (this is 1^st positive accel peak) t = 3/16 = 0.1875 seconds		(d) s = A (4 / T ) * t s = 6cm(4 / (1/8 sec)) 1sec s = 192 cm
		The wave travels one amplitude every quarter of a period (T/4) · (A) 0 amplitude to peak (T/4) · (B) peak back to 0 · (C) 0 to trough · (D) trough back to 0

Ch 15.3 #19

A 50.0 g object connected to a spring with a force constant of 35.0 N/m oscillates on a horizontal, frictionless surface with amplitude of 4.00 cm. Find (a) the total energy of the system and (b) the speed of the object when the position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when the position is 3.00 cm. ans: 28 mJ b. 1.02 m/s c. 12.2 mJ d. 15.8 mJ

ω^2 = k/m ω = 26.46 rad/s x(t) = A cos(ωt) 0.01 = 0.04cos(26.46t) t = 0.050 sec 0.03 = 0.04cos(26.46t) t = 0.0273 sec

(a) E = ½kA^2 E = ½35(0.04)^2 E = 0.028 Joules = 28 mJ (b) v = -A ω sin(ωt) v = -.04ωsin(26.46*.05) |v| = 1.02 m/s

(c) K = ½kA^2sin^2(ωt) K = ½35(.04)^2sin^2(26.46*.0273) K = .01225 Joules = 12.3 mJoules (d) U = ½kA^2cos^2(ωt) U = ½35(.04)^2cos^2(26.46*.0273) U = 15.8 mJ

Ch 15.5 pendulum A 1.00 kg pendulum with a length of 2.00 m is released at an angle of 10.0° from the vertical. What is the (a) speed at the bottom and (b) the restoring force at the extremes by using the simple harmonic motion model? (c) Solve this problem more precisely by using more general principles of Newton’s laws (Ch 6).

A = r q A = 2m (10° (p/180°) A = 0.349 m		ω^2 = g / L ω^2 = 9.8 / 2 ω^2 = 4.90 rad/sec	(a) vmax = A ω vmax = 0.349 * 4.9 vmax = 0.820 m/s		(b) Fmax = m amax Fmax = m A ω^2 Fmax = 1 (.349)(4.9^2) Fmax = 8.38 N
(d)	h = r - cosqr h = 0.0341 ½mv^2 = mgh v = 0.8176 m/s			F = ma = mgsinq a = 2.54 m/s^2 a = a r a = 2.54 m/s^2		F = mgsinq F = 0.635 N

Ch 15.6 #40

Show that the time rate of change of mechanical energy for a damped, undriven oscillator is given by dE/dt = –bv² and hence is always negative. Proceed as follows: Differentiate the expression for the mechanical energy of an oscillator, E = ½mv² + ½kx², and use equation ma = -kx – bv, where a = d²x/dt².

TE = ½ m v^2 + ½ k x^2 take time-derivative dE/dt = mv d^2x/dt^2 + kx dx/dt dE/dt = v m d^2x/dt^2 + kx v (Given: m d^2x/dt^2 = -kx – bv)

dE/dt = v(-kx – bv) + kx v dE/dt = -vkx – bv^2 + kx v dE/dt = – bv^2 < 0

Ch 15.7 #48

Damping is negligible for a 0.150 kg object hanging from a light 6.30 N/m spring. A sinusoidal force with amplitude of 1.70 N drives the system. At what frequency will the force make the object vibrated with amplitude of 0.440 m? 1.31Hz or 0.641 Hz

; ωo^2 = k/m When b is neglible the above equation simplifies down to ω^2 = k/m ± Fo /(m*A)

A = (Fo / m) / ± (ω^2 - ωo^2) ω^2 = ωo^2 ± Fo / (m*A) ω^2 = k/m ± Fo /(m*A) ω^2 = 6.3/.15 ± 1.7/(.15*.44) ω = 8.23 or 4.03 rad/sec ω = 2πf f = 1.31 Hz or 0.641 Hz