Question

<HW 6 (partial) Problem 14.61 A 150 g ball attached to a spring with spring constant 2.20 N/m oscillates horizontally on a frictionless table. Its velocity is 18.0 cm/s when = -5.00 cm. Part A What is the amplitude of oscillation? Express your answer in centimeters. % AED O 2 ? A = Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part B What is the speed of the ball when = 3.00 cm? Express your answer in centimeters per second. IVO AED 0 2 ? cm/s Submit Request Answer < Return to Assignment Provide Feedback

I have tried 7.38 cm for the part A, which was incorrect.

Answer 1

The angular fouequency of osseillation is work Here K= Sprung constant = 2.20N/m M = mass = 1509 = 0.150kg W W = Vo. 150kg = 2.20 N/m = $14.67 W = 3.8305! The equation of oscillator motion is a(t) = A.sin(wt+%) The speed of the oscillator is, V(t) = W.A.COS(Wt+Ø) 0 Theirefore, X(t) | 0(H) = A. Sinccrto) CO.A. COS(tot18) act) = tan(wt+6) - V(t) tan(wt+º) = w.act) ut)

Here act) = -5.00cm (t)= 18.0cm/s taniwt+9) = 3.830 5x -5.000m 18.00mis = 3.830 X-0.278 tanculto) = -1.06474 wtt8 = tan 61.06474) Wtto = -46.796 degrees.

PortA: -Amplitude A is, A = act) Sinct+) = -5.00cm sin(-46.7960) [from ano] = -5.00cm -0.7289 A = 6.86cm 1 Amplitude (A) = 6.86cm! Path:- when acti)= 3 m we have 3 = 6.86 x Sin(whitØ) (From cho 3cm Sinfotito) = 6.86 cm = 0.4373 Cuti +9 = sin(0.4373) Wtito = 25.90 WCh) = W.A.cos (ct1+0) = 3.830 8'x 6.86cm X COS (25.93) = 26.2738 X 0.8993 cmis = 23.62cm/s 2 = 23.62 omis