Question

A mass m = 2.77 kg is attached to a spring of force constant k = 44.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.11 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle has a constant counterclockwise angular speed w, constant tangential speed V = WA, and centripetal acceleration of constant magnitude a = w?A. X-A +A ΤωΑ = 'A

Answer 1

bampe X0 നാണ്, 22/+A 16-WA X2 W А 6 Starts when n=tA V=O x = conut A wty) x X(t) = A Concet 44.9 W/m m ka Spring Constant - - 2-77kg m = mass 4.03 rad/sec. A Amplitude. - 11m X(t) Il Cop(4.03€) Now differentiateco X(t) with res pact to t. duct) Ad Closwt) dt dt = A (-w) sinct) - - WA sinut 111 rct) =relocity relt) - (4.03) (+11) Sin (4.03t) -.44 sin(4.03t) 1 I dy drct) - ult) at Now differentiate ult) with respect to t. We get acceleration = a. druct) - WA.d (sinut) at dt - WA X W Cosat = - W2 A boswt a. a = -( 4. 03) x (11) Ca A (4* 0 3+) -- | 79 Co (4 -13 )

Ans V = 11 cos(4.03€) - 44 sin(4.03t) - 1.79 GA (4.03+) a When starts at x=0, v=twa X(t) is vertical component of displacement. xet) = Asinat:.... A 26 Twit we get Now differentiate. &(t) with respect to t velocity = re. du at. Ard (sinut) dt Aw conwt v= we get twith respect tot Now differentiate re acceleration, dre at die at Aw d (conut) at - Aw. (W) Sin wt -wra sinut Am W- 4.03 rad/sec, A2.11 m. X(t) oll sin(4.03€) r = (-11) (4.03) Con (4.03t) 44 Con (4.03+) - (4.03) 2 (-11) solo con sin (4:03+) -1.79 sin (4.03t).