12 homework RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am
Gravity
^ F21 = -G m12m2 r12 ,
r12
1
for r R,
g(r) = G M r2
G = 6.67259 10-11 N m2 /kg2 Rearth = 6370 km, Mearth = 5.98 1024 kg
2 Circular orbit: ac = v = 2 r = r
2 2 r = g(r) T M U = -G mr , E = U + K = - G mrM 2 2 F = - d U = -m G M = -m v 2 r dr r r0 r2 = 1- = 2L = const. m
1 Intensity: I = P = 2 v ( smax )2 A I Intensity level: = 10 log10 I , I0 = 10-12 W/m2 0 Plane waves: (x, t) = c sin(k x - t) c Circular waves: (r, t) = sin(k r - t) 1 v Doppler effect: = v T , f0 = T , f = Here v = vsound vobserver , is wave speed relative to moving observer and = (vsound vsource )/f0 , detected wave length established by moving source of frequency f0 . freceived = fref lected c Spherical: (r, t) = r sin(k r - t) r
s P = -B V = -B x V Pmax = B smax = v smax x Piston: m = m At = A v t V
Kepler's Laws of planetary motion: r0 0 i) elliptical orbit, r = 1- rcos r1 = 1+ ,
1 ii) L = r m r - A = 2 r r t t t
21 r1 +r2 , T 2 = 4 2 r 3 iii) G M = 2 T a a, a = 2 GM a2 Escape kinetic energy: E = K + U (R) = 0
Fluid mechanics
Pascal: P = F1 = F2 , 1 atm = 1.013 105 N/m2 A1 A2 Archimedes: B = M g, Pascal=N/m2 P = Patm + g h, with P = F and = m A V
1 Shock waves: Mach Number= vsource = sin vsound
Superposition of waves
h F = P dA - g 0 (h - y) dy Continuity equation: A v = constant Bernoulli: P + 1 v 2 + g y = const, 2
P 0
Phase difference: sin(k x - t) + sin(k x - t - ) Standing waves: sin(k x - t) + sin(k x + t) Beats: sin(kx - 1 t) + sin(k x - 2 t) Fundamental modes: Sketch wave patterns String: = , Rod clamped middle: = , 2 2 Open-open pipe: = , 2
Oscillation motion
1 f = T , = 2 T 2 2 x S H M: a = d 2 = - 2 x, = d 2 = - 2 dt dt x = xmax cos( t + ), xmax = A v = -vmax sin( t + ), vmax = A a = -amax cos( t + ) = - 2 x, amax = 2 A E = K + U = Kmax = 1 m ( A)2 = Umax = 1 k A2 2 2 Spring: m a = -k x Simple pendulum: m a = m = -m g sin Physical pendulum: = I = -m g d sin Torsion pendulum: = I = -
Open-closed pipe: = 4
Temperature and heat
Conversions: F = 9 C + 32 , K = C + 273.15 5 Constant volume gas thermometer: T = a P + b 1 Thermal expansion: = 1 d , = V d V dT dT = T , A = 2 A T , V = 3 V T Ideal gas law: P V = nRT = N kT R = 8.314510 J/mol/K = 0.0821 L atm/mol/K k = 1.38 10-23 J/K, NA = 6.02 1023 , 1 cal=4.19 J Calorimetry: Q = c m T, Q = L m First law: U = Q - W , W = P dV Conduction: H = Q = -k A T , Ti = -H ki t A i W Stefan's law: P = A e T 4 , = 5.67 10-8 m2 K 4
Wave motion
Traveling waves: y = f (x - v t), y = f (x + v t) In the positive x direction: y = A sin(k x - t - ) 1 T = f , = 2 , k = 2 , v = = T T k
F fixed end: phase inversion Reflection of wave: open end: same phase General: E = K + U = Kmax P = E = 1 m (A)2 2 t t Waves: m = m x = m v t x t x 1 P = 2 v ( A)2 , with = m x Circular: m = m A r = m 2 r v t A r dt A Spherical: m = m 4 r2 v t V
Kinetic theory of gas
Along a string: v =
v F = px = md x t N 2 Pressure: P = N F = m N vx = mV v 2 A V 3 1 P = 2 N K, K x = K = 2 k T , T = 273 + Tc , 3 V 3 P V = N k T , n = N/NA , k = 1.3810-23 J/K, NA = 6.02214199 1023 #/kg/mole Constant V: Q = U = n CV T Constant P: Q = n CP T
Ideal gas: px = 2 m vx ,
2
Sound
v=
B,
s = smax cos(k x - t - )
= C P , CP - C V = R V CV = d R, for transl.+rot+vib, d = 3 + 2 + 2 2 Adiabatic expansion: P V = constant 1 Mean free path: = (v vrms t t d2 n = 1 2 )
rel rms V
C
2 d nV
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am
2
Mechanics - Basic Physical Concepts
Math: Circle: 2 r, r 2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = -b 2 a-4 a c
Cartesian and polar coordinates: y x = r cos , y = r sin , r 2 = x2 + y 2 , tan = x Trigonometry: cos cos + sin sin = cos( - ) sin + sin = 2 sin + cos - 2 2 cos + cos = 2 cos + cos - 2 2 sin 2 = 2 sin cos , cos 2 = cos2 - sin2 1 - cos = 2 sin2 2 , 1 + cos = 2 cos2 2 Vector algebra: A = (Ax , Ay ) = Ax^ + Ay i ^ Resultant: R = A + B = (Ax + Bx , Ay + By ) Dot: A B = A B cos = Ax Bx + Ay By + Az Bz Cross product: ^ = k, k = ^, k ^ = i ^ ^ ^ ^ i ^ i ^ ^ i Ax Bx ^ Ay By ^ k Az Bz
B F s = F s cos = F s A F ds
(in Joules)
Effects due to work done: Fext = m a - Fc - fnc Wext |AB = KB - KA + UB - UA + Wdiss |AB B Kinetic energy: KB -KA = A m ads, K = 1 m v 2 2
B K (conservative F ): UB - UA = - A F ds 1 k x2 Uspring = 2 Ugravity = m g y, From U to F : Fx = - U , Fy = - U , Fz = - U x y z
Fspring = - U = -k x x U = 0, 2 U > 0 stable, < 0 unstable Equilibrium: x x2 W Power: P = ddt = F v = F v cos = F v (Watts)
Fgravity = - U = -m g, y
Collision
C =AB =
d d 1 Calculus: dx xn = n xn-1 , dx ln x = x , d sin = cos , d cos = - sin , d d d dx const = 0
C = A B sin = A B = A B , use right hand rule
Measurements
Impulse: I = p = pf - pi tif F dt Momentum: p = m v x Two-body: xcm = m1m1 +m2 x2 1 +m2 pcm M vcm = p1 + p2 = m1 v1 + m2 v2 Fcm F1 + F2 = m1 a1 + m2 a2 = M acm K1 + K2 = K1 + K2 + Kcm Two-body collision: pi = pf = (m1 + m2 ) vcm vi = vi - vcm , vi = vi + vcm Elastic: v1 - v2 = -(v1 - v2 ), vi = -vi , vi = 2 vcm - vi Many body center of mass: rcm = Force on cm: Fext = dp = M acm , dt
mi r i = mi r dm mi
t
Dimensional analysis: e.g., 2 F = m a [M ][L][T ]-2 , or F = m v [M ][L][T ]-2 r N (a x + b) = a N x + b N Summation: i i=1 i=1 i
p=
pi
Rotation of Rigid-Body
Motion
One dimensional motion: v = d s , a = d v dt dt s -s v -v Average values: v = tf -tii , a = tf -tii f f One dimensional motion (constant acceleration): v(t) : v = v0 + a t +v s(t) : s = v t = v0 t + 1 a t2 , v = v02 2 2 = v2 + 2 a s v(s) : v 0 Nonuniform acceleration: x = x0 + v0 t + 1 a t2 + 2 1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .) 6 24 120 720 Projectile motion: trise = tf all =
ttrip v0y 2 = g 1 g t2 , R = v t h = 2 f all ox trip 2 1 Circular: ac = v , v = 2 T r , f = T (Hertz=s-1 ) r 2 + a2 Curvilinear motion: a = at r
Isphere = 2 M R2 , Ispherical shell = 2 M R2 5 3 I = M (Radius of gyration)2 , I = Icm + M D2 Kinetic energies: Krot = 1 I 2 , K = Krot + Kcm 2 Angular momentum: L = r m v = r m r = I Torque: = d L = m d v r = F r = I d = I dt dt dt Wext = K +U +Wf , K = Krot + 1 m v 2 , 2 P =
s Kinematics: = r , = v , = at r r 2 Moment of inertia: I = mi ri = r2 dm 2 2 Idisk = 1 M R2 , Iring = 1 M (R1 + R2 ) 2 2 1 M 2, I 1 M (a2 + b2 ) Irod = 12 rectangle = 12
Rolling, angular momentum and torque
I Rolling: K = 1 Ic + M R2 2 = 1 Rc + M v 2 2 2 2 Angular momentum: L = r p, L = r p = I
Relative velocity: v = v + u
Law of Motion and applications
Force: F = m a, Fg = m g, F12 = -F21
2 Circular motion: ac = v , v = 2 T r = 2 r f r Friction: Fstatic s N Fkinetic = k N
Torque: = d L = r d p = r F , = r F = I dt dt 1 Gyroscope: p = d = L d L = L = m g h I dt dt
Static equilibrium
Equilibrium (concurrent forces):
i Fi = 0
Energy
Work (for all F): W = WAB = WB - WA
i = 0 Fi = 0, about any point Subdivisions: rcm = mA rAcm +mB rBcm mA +mB Elastic modulus = stress/strain stress: F/A strain: L/L, x/h, -V /V
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am Version number encoded for clicker entry: V1:1, V2:4, V3:3, V4:5, V5:2. Question 1 part 1 of 1 10 points Earthquakes produce two kinds of seismic waves: The longitudinal primary waves (called P waves) and the transverse secondary waves (called S waves). Both S waves and P waves travel through Earth's crust and mantle, but at different speeds; the P waves are always faster than the S waves, but their exact speeds depend on depth and location. For the purpose of this exercise, we assume the P wave's speed to be vP = 9620 m/s while the S waves travel at a slower speed of vS = 4000 m/s. Suppose a seismic station detects a P wave and then t = 83.3 s later detects an S wave. How far away is the earthquake center? Correct answer: 570.353 km (tolerance 1 %). Explanation: Suppose the earthquake happens at time t = 0 at some distance d. The P wave and the S wave are both emitted at the same time t = 0, but they arrive at different times, respectively tP = d/vP and tS = d/vS . The S wave is slower, so it arrives later than the P wave, the time difference being t = d d(vP - vS ) d - = . vS vP vP vS
3
where A = 0.59 m, = 8.4 s-1 , k = 5.1 m-1 , = 0.3, t is in seconds and x and y are in meters. What is the speed of the wave? Correct answer: 1.64706 m/s (tolerance 1 %). Explanation: Basic Concepts: v= k= T
2 f= 2 Solution: For the propagation speed of a wave we have v= k (8.4 s-1 ) = (5.1 m-1 ) = 1.64706 m/s .
Question 3 part 2 of 5 10 points What is the vertical displacement of the string at x0 = 0.074 m, t0 = 0.9 s? Correct answer: 0.549778 m (tolerance 1 %). Explanation: We can obtain the result if we simply substitute the values for x and t into y(x, t) y(x0 , t0 ) = A sin[ t - k x + ] = (0.59 m) sin (8.4 s-1 ) (0.9 s) - (5.1 m-1 ) (0.074 m) + (0.3) = 0.549778 m .
Consequently, given this time difference and the two waves' speeds vP and vS , we find the earthquake center being d= vP vS t = 570.353 km vP - v S
away from the seismic station. Question 2 part 1 of 5 10 points The wave function for a linearly polarized wave on a taut string is y(x, t) = A sin( t - k x + ),
Question 4 part 3 of 5 10 points What is the wavelength? Correct answer: 1.232 m (tolerance 1 %).
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am Explanation: For the wave vector we have k= Thus 2 = k 2 . Question 7 part 1 of 2 10 points
4
A wave moving along the x axis is described by y(x, t) = A e-
(x+v t)2 b
2 = (5.1 m-1 ) = 1.232 m .
where x is in meters and t is in seconds. Given A = 3 m, v = 5 m/s, and b = 5 m2 . Determine the speed of the wave. Correct answer: 5 m/s (tolerance 1 %). Explanation: y(x, t) = A e-
(x+v t)2 b
Question 5 part 4 of 5 10 points What is the frequency of the wave? Correct answer: 1.3369 Hz (tolerance 1 %). Explanation: Since = 2 f , we have f= 2 (8.4 s-1 ) = 2 = 1.3369 Hz .
is of the form
y(x, t) = f (x + v t) so it describes a wave moving in the negative x direction at v = 5 m/s .
Question part 8 2 of 2 10 points Determine the direction of the wave.
Question 6 part 5 of 5 10 points What is the maximum magnitude of the transverse speed of the string? Correct answer: 4.956 m/s (tolerance 1 %). Explanation: Taking derivative of y over t, we find magnitude of the transverse speed of the string: y(x, t) vy = t = A cos( t - k x + ) . Thus vy,max assumes its maximum value when t - k x + = 2 n, n = 0, 1, 2, ...; i.e., when cos( t - k x + ) = 1, so vy,max = A = (0.59 m) (8.4 s-1 ) = 4.956 m/s .
1. +x direction 2. -x direction correct 3. +y direction 4. -y direction Explanation: Question 9 part 1 of 1 10 points A wave is sent along a long spring by moving the left end rapidly to the right and keeping it there. The figure shows the wave pulse at QR - part RS of the long spring is as yet undisturbed. P S
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am P Q R S x 7. Which of the graphs correctly shows the relation between original position x of each point of the spring and the corresponding displacement x of that point? (Displacements to the right are positive.) x 1. x x
5
x 2. x correct x 3. x
Explanation: The left end of the long spring (point P) is displaced towards the right and kept there. The displacement x of point P is therefore nonzero (positive). Only two graphs show a nonzero displacement at the left end. At the instant shown, the wave pulse has not yet reached the right end of the long spring (point S), so this end has not yet been displaced. One of the two graphs shows a negative displacement at the right end, so the only possible choice is the other one. x
x Question 10 part 1 of 1 10 points A harmonic wave
x 4.
x
y = A sin[k x - t - ] , where A = 1 meter, k has units of m-1 , has units of s-1 , and has units of radians, is plotted in the diagram below. At the time t = 0 +1 A (meters) -1 x (meters) 6 12 18 Which wave function corresponds best to the diagram?
5. None of these graphs are correct. x 6. x
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am 2 1 x-t- 15 m 3 2 2 x-t- 3m 3 2 5 x-t- 3m 3 2 2 x- t- cor9m 3 2 x-t- 9m 2 x-t- 15 m 2 x-t- 15 m 2 x-t- 9m 2 x-t- 3m 2 x-t- 15 m 5 3 5 3 4 3 4 3 4 3 2 3 (gray curve in diagram below), therefore =2 2 = radians . 3 3
6
1. y = A sin 2. y = A sin 3. y = A sin 4. y = A sin rect
Checking the wave function y at x = 0, we have agreement with the diagram below y = sin 0 - 2 3 = -0.866025 m .
At the time t = 0 +1 A (meters) -1 x (meters) 9 =9 18 Therefore, the wave function is y = A sin 2 9m x-t- 2 3 . keywords: Question 11 part 1 of 2 10 points Tension is maintained in a string as in the figure. The observed wave speed is 22 m/s when the suspended mass is 3.1 kg . The acceleration of gravity is 9.8 m/s2 .
5. y = A sin 6. y = A sin 7. y = A sin 8. y = A sin 9. y = A sin 10. y = A sin
Explanation: From the diagram of the wave function the wave-length = 9 m (6 horizontal scale divisions of 1.5 m each, see diagram below). Notice: Since one wave-length is 2 ra 2 = dians, each horizontal division is 6 3 radians. The given wave function (sine function with t = 0) y = A sin k x - = A sin = A sin 2 x- 2 x- 9m
(dark curve in diagram below) is shifted 2 divisions to the right (negative phase shift) of a no-phase-shift sine function y = sin 2 9m 3.1 kg x
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am What is the mass per unit length of the string? Correct answer: 0.0627686 kg/m (tolerance 1 %). Explanation: Let : v = 22 m/s and m = 3.1 kg . The tension in the string is F = mg and its velocity is v= F , and Question 13 part 1 of 1 10 points
7
You are given f1 (x), a transverse wave that moves on a string that ends and is FIXED in place at x = 5 m. As the problem begins, the wave is moving to the right at v = 1 m/s. Amplitude (centimeter) 3 2 1 0 -1 -2 -3 0 1 2 v
F = 2 v mg = 2 v (3.1 kg) 9.8 m/s2 = (22 m/s)2 = 0.0627686 kg/m . Question 12 part 2 of 2 10 points What is the wave speed when the suspended mass is 2.5 kg? Correct answer: 19.7566 m/s (tolerance 1 %). Explanation: Let : The velocity is v= = = F mg (2.5 kg) (9.8 m/s2 ) 0.0627686 kg/m m = 2.5 kg .
3 4 5 6 7 8 9 10 Distance (meter) Hint: Consider the image of the wave reflected about the FIXED point x = 5 m in the following diagram. The image will be moving to the left at v = -1 m/s (in the opposite direction from the real wave). Amplitude (centimeter) 3 2 1 0 -1 -2 -3 0 1 2 v v
3 4 5 6 7 8 9 10 Distance (meter) What is the shape of the wave on the string after 3 s?
= 19.7566 m/s .
3 2 1 1. 0 -1 -2 -3 0 1
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am 3 2 1 5. 0 -1 -2 2 3 4 5 6 7 8 Distance (meter) 9 10 -3 0 1 2 3 4 5 6 7 8 Distance (meter)
8
9 10
3 2 1 2. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
correct 3 2 1 6. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
3 2 1 3. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
3 2 1 7. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
3 2 1 4. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
3 2 1 8. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am 3 1 9. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10 Amplitude (centimeter) 2 3 2 1 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) Superposition, at t = 3 s
9
9 10
3 2 1 10. 0 -1 -2 -3 0 1 2 3 4 5 6 7 8 Distance (meter) 9 10 Amplitude (centimeter) 3 2 1 0 -1 -2 -3 0 1 2
Resultant, at t = 3 s
3 4 5 6 7 8 Distance (meter)
9 10
Question 14 part 1 of 2 10 points Explanation: The initial wave (real) on the string is represented with a dashed line and its reflected wave (imaginary) is represented with a dotted line. Initial time, t = 0 s 3 Amplitude (centimeter) 2 1 0 -1 -2 -3 0 1 2 A sinusoidal wave of the form
y = A sin(k x - t)
is traveling along a string in the x direction, where A = 0.83 mm , k = 1.5 m-1 , = 33 rad/s , with x in meters and t in seconds. For this string, the mass per unit length is given by = 0.01 kg/m. 0.83 mm 0.19 s 0.19 s -0.83 mm
3 4 5 6 7 8 9 10 Distance (meter) After 3 s the positions of the two waves are have both moved 3 meters in opposite directions. The resultant sum of the two waves is the light gray line.
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am 0.83 mm
m
10
4.19 m 4.19 m -0.83 mm What is the maximum value of the acceleration of the segment of the string at x = 0.5 m? Correct answer: 0.90387 m/s2 (tolerance 1 %). Explanation: The maximum acceleration is independent of x. amax = d2 y dt2
2 (33 rad/s) = 0.1904 s , and 2 = k 2 = (1.5 m-1 ) = 4.18879 m . = Question 16 part 1 of 3 10 points A wave pulse traveling along a string of linear mass density 0.0044 kg/m is described by the relationship y = A0 e-b x sin(k x - t) , where A0 = 0.0084 m, b = 0.61 m-1 , k = 0.37 m-1 and = 19 s-1 . What is the power carried by this wave at the point x = 1.5 m? Correct answer: 0.000461617 W (tolerance 1 %). Explanation: We know for a sinusoidal wave the power transmitted in it is P = 1 2 A2 v . 2
max
= A 2 = (0.83 mm) (33 rad/s)2 = 0.90387 m/s2 . Question 15 part 2 of 2 10 points Hint: You may work on this part without the answer to the first part. For a length segment x = 1 cm along the string, what is the total energy of oscillation? Correct answer: 3.75106 10-8 J (tolerance 1 %). Explanation: The total energy is E = K + U = Kmax 1 2 = m vmax 2 1 = x ( A)2 2 = 3.75106 10-8 J .
Here A = A0 e-b x and v = /k, then we have P = 1 3 A2 e-2 b x 0 2k (0.0044 kg/m)(19 s-1 )3 = (0.0084 m)2 -1 ) (2)(0.37 m e-[2 (0.61 m ) (1.5 m)] = 0.000461617 W .
-1
Question 17 part 2 of 3 10 points What is the power, P0 , carried by this wave at the origin? Correct answer: 0.00287767 W (tolerance 1 %). Explanation:
Note : T =
2
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am At the origin x = 0, we have P0 = 1 3 A2 0 2k (0.0044 kg/m) (19 s-1 )3 (0.0084 m)2 = 2 (0.37 m-1 ) = 0.00287767 W .
11
Correct answer: 3.26513 kg (tolerance 1 %). Explanation: From the free body analysis of the vertical forces, we have m g = 2 T sin , mg T = . 2 sin The angle can be found from 3L cos = 8 L 2 = 0.75 , so = arccos(0.75) = 41.4096 . Using Eq. 1, the wave speed is v= = T so (1)
Question 18 part 3 of 3 10 points P Compute the ratio . P0 Correct answer: 0.160414 (tolerance 1 %). Explanation: The ratio of them is R = e-2 b x -1 = e-2(0.61 m ) (1.5 m) = 0.160414 .
Question 19 part 1 of 1 10 points A light string of mass per unit length 5.69 g/m has its ends tied to two walls separated by a distance equal to three fourths the length L of the string as shown in the figure. A mass m is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9.8 m/s2 . 3L 4
mg , so 2 sin sin m = 2 v2 g = 2 (65.2 m/s)2 (3.26513 kg) sin(41.4096 ) (9.8 m/s2 ) = 3.26513 kg . Question 20 part 1 of 2 10 points Assume: The mass density and all other properties of the rope and wave are unchanged except those considered below. A transverse wave is being generated on a rope under constant tension. The power transmitted by the wave is increased or decreased by what factor if the rope is replaced by an identical rope twice as long? 1. P = 1 P0 2
L 2
L 2
mg What size mass should be suspended from the string to produce a wave speed of 65.2 m/s?
2. P = 2 P0
homework 12 RAMSEY, TAYLOR Due: Nov 20 2006, 4:00 am 1 3. P = P0 4 4. P = 4 P0 5. P = P0 correct Explanation: Basic Concepts: P = 1 v 2 A2 2 F From (1) and (2) it follows that P 2 A2 .
12
(3)
Thus, if A is doubled and is halved, P 2 A2 remains constant.
v=
v For all points, the force F is constant, = 2 P = 1 v 2 A2 2 F , (1)
v=
(2)
it follows that if L is doubled, v remains constant and P is constant. Question 21 part 2 of 2 10 points Assume: All other properties of the rope and wave are unchanged except those considered below. The power transmitted by the wave is increased or decreased by what factor if the amplitude of the original wave is doubled and the angular frequency is halved? 1. P = 1 P0 2
2. P = 2 P0 3. P = 1 P0 4
4. P = 4 P0 5. P = P0 correct Explanation: