Question

Given:

A(amplitude)=6.13 cm; x(0)= -5.4 cm; T(period) = 1.7s; a(0)=73.8 cm/s^2; v(0)= 10.7 cm/s

I need a solution to the following problem:

Part D What is the phase,expressed in degrees, at each of three different times: the first time is one-third of a period, t 0.5667 S;the second time is two-thirds of a period,t 1.133 s; and the third time is one full period after the clock starts, t 1.70 s? Give your answers for the phase in degrees. Give your answers as an ordered triple, with the phase at 0.5667 s first, followed by a comma, followed by the phase at 1.133 s, followed by a comma, followed by the phase at 1.70 s.Give each of your answers as an angle in degrees at 0.5667 s, at 1.133 s, at 1.70 s-31.8,88.2,208 degrees, degrees, degrees

As you can see I already requested the answer so I don't know how to solve it. Also what do they mean by 'phase'? i thought the phase constant is constant so this is confusing.

Answer 1

aqi ne ver Time period (T) 1.4 Secondu The objecve is to determine the phase Now ωe calculate 3.695 x &o At time todisplace ment x-54 C A vahuo Substitute all values and ω 103.695A-29.16

c8e Squaring on botk sides As 8.38A- 2.16 A tie Amplituda, A-6 . 1 2 cm uwe equation A Sin (wt) At t-o, -5.4= 6.12 Sin (RS O.2Sin Sin 1(-0.8823 on value inwae So h subsibutingV wilbecome t.-61.92° Sin (3.695 χ-A 6 12 Sin(it. 6 t-61.92 at Wow we determine phae t: o. 566 Sec i: L6.12 Sin( 211.96 x o. 566)-61.92 phase 6.12 Sin (i19.856) -61.92 = 6, 12. ХО.86 2-61.92. 5.30- 61.92 。 =-56.612

net 3 At t133 Se Sin(ขเรเท.is)-61.92) orgs phase skil (6.12 Sin (239.42)- G192 6.12(-o.?6 53)-61.92) 5.296-61.92 2616 phaje au 6 12 Sin 359.9-61.92 6-12 (-1.39 ‘xor)-61.92 -8.545 x6.92 3 -61.920°