CALC ?(a) Show that for a wave on a string, the kinetic energy per unit length of string is where is the mass per unit length. (b) Calculate for a sinusoidal wave given by Eq. (15.7). (c) There is also elastic potential energy in the string, associated with the work required to deform and stretch the string. Consider a short segment of string at position x that has unstretched length ?x as in Fig. 15.13. Ignoring the (small) curvature of the segment, its slope is Assume that the displacement of the string from equilibrium is small, so that has a magnitude much less than unity. Show that the stretched length of the segment is approximately (Hint: Use the relationship valid for |u|«1.) (d) The potential energy stored in the segment equals the work done by the string tension F (which acts along the string) to stretch the segment from its unstretched length ?x to the length calculated in part (c). Calculate this work and show that the potential energy per unit length of string is (e) Calculate are maximum where y is zero, and vice versa. (h) Show that the instantaneous power in the wave, given by Eq. (15.22), is equal to the total energy per unit length multiplied by the wave speed Explain why this result is reasonable.
Trigonometry exam 3 study guide Pythagorean Identities: 2 2 Sin θ+Cos θ=1 2 2 Tan θ+1=Sec θ 2 2 1+Cot θ=Csc θ X 0 π π π π Conjugates: 6 4 3 2 A+B--------A-B so (a+b)*(a-b)=a -b Sin 0 1 √ 2 √3 1 2 2 2 sin α−sin α=cos α−cos α 4 Co 1 3 2 1 0 √ √ s 2 2 2 sin α+cos α ≠ 1 Ta 0 √ 3 1 √ 3 DN 2 2 Instead: sin α=(1−cos α) and n 3 E sin α=(1−cos α)(1−cos α) or 1−2cos α+cos α4 Sin + All + Csc + Tan + Cos + Cot + Sec + 1 cosθ= 2 Cosine is positive in quadrants 1 and 4 and has two solutions 1 π cosr= 2sr= 3 Use the table to find the value of the reference angles Q1: π Q4: 5π 3 3 Assign the values from the unit circle Q1:θ= +k∗2π 3 Add two rotations to equal θ Q4:θ= 5π +k∗2π 3 Answer for sin and cos problems must have two values Let’s try a tangent problem: tanθ= √3 3 tanrisr= π 6 Q1:θ= +k∗π 6 tangent problems only require one value because tangent problems create a straight diagonal line Now for a more interesting problem: −2sin3θ+1=0 1 sin3θ= Use algebra to create a rational equation (note: sin3θ ≠ 3sinθ) 2 sinr= isr= π 2 6 Q1:3θ= +k∗2π 6 Write answers in terms of 3θ Q2:3θ= 5π +k∗2π 6 Q1:θ= π + k∗2π 18 3 Use division Q2:θ= 5π +∗2π 18 3 Next something even harder: 2 4 csc θ= 3 2 cscθ=+¿− Find the square roots √3 √3 sinθ=+¿− Reciprocate for a sin problem 2 √3 π sinr= isr= 2 3 π Q1:θ= +k∗2π There should be an answer in every quadrant 3 2π Q2:θ= 3 +k∗2π because the reference angle is positive and negative 4π Q3:θ= 3 +k∗2π 5π Q4:θ= +k∗2π 3 Now for Quadratic equations: 2 cos θ−cosθ=0 cosθ(cosθ−1 )=0 Factor cosθ=0∨cosθ=1 cosr=1isr=0 π cosr=0isr= 2 π π θ= +k∗π 2 Quadrants are irrel2van and 0 represent a right θ=k∗2π angle Last one: cosθ=1−sin 2 1−sinθ¿ cosθ¿ =¿ Square both sides ¿ cos θ=1−2sinθ+sin θ 2 1−sin θ=1−2sinθ+si n θ 2 Use Pythagorean identities 2sin θ−2sinθ=0 Move terms to one side 2sinθ(sinθ−1=0 Factor 2sinr=0issinr=0isr=0 π sinr=1isr= 2 θ= +k∗π Remember to check for extraneous solutions 2 θ=k∗2 π Cos(a-b) = cosacosb + sinasinb Cos(a+b) = cosacosb – sinasinb Sin(a-b) = sinacosb - cosasinb Sin(a+b) = sinacosb + cosasinb tana−tanb Tan(a-b) =1+tanatanb tana+tanb Tan(a+b) = 1−tanatanb Here is an example problem: 7π 7π 2π π Find the Exact value ofn12 (hint: 12 = 3 − 4 ) sin 2π− π =sin2π cosπ −cos2π sin (3 4 ) 3 4 3 4 Plug in the values ¿ √3 √2 − −1 √2 Derive reference ( 2 ( 2 (2) (2) angles 6 − 2 ¿ √ − √ Multiply 4 4 6+ 2 √ √ Combine (this is the 4 finished form of the problem Double angle formulas: Half angle formulas(+ and – is either/ or not both): Sin: sin2x=2sinxcosx Sin: x 1−cosx sin2=± √ 2 2 2 Cos: cos2x=cos x−si n x Cos: x 1+cosx cos2=± √ 2 2 cos2x=2cos x−1 Tan: tan = 1−cosx 2 sinx cos2x=1−2si n x tan2 x=2tanx Tan: 1−tan x −29 3π −21 −20 Example: secθ= π