Question

QUESTION 23 A block (m=20g) is attached to a spring, pulled 10 cm to the right, and released at t-0. It makes 12 complete oscillations in 10 s. Assume initial phase angle -0 in the oscillation equations, (b) What is its position x, velocity , and acceleration a att-2 s? Simple harmonic motion of block Ar -0.0966 m; -0.1962 m/s; = -5.489 m/s? B. = -0.0966 m; y= -0.1962 m/s; - +5.489 m/s C -0.8092 m; 4.4297 m/s; y +46 m/s D y a -0.08092 m; +0.44297 m/s; +4.60 m/s? -0.08092 m; E -0.44297 m/s; = +4.60 m/s

Answer 1

The.equation of simple harmonia metion is ..R . a cos (we - $), ---- (1). where a → displacement, a → amplitude, ..^ => angular frequency, $ + ventiset phase angle. Velocity of the block, voda veraw Sin (wt - ) --(2) .Aueleration of the block., A dv. A = - a wé cos (wt - 4) ---(3) In this problem, initial phase angle 0-0. .. a equations (1), (2) and (3) becomes...

X = a (swt) Ve saw Sin(wt) A -aw Coswt ) let us use the values given in the problem, . Number of osullations in 10 seconds 12 e Timeperiod of one oscillation which is defined as the time taken to complete one osculation, ... I, 10 0.833 S . . We know We 20 21 Amplitude is given, At t.29, aa Loom 1.X10.cos ( 21 x 2) [ taking expo (4] X = - 8,093 cm .......... . - 0.0%093 m () Taking equation (5) for Nelocity. .. v - aw Sin (wt) 2.-10x21. Sin (2002) 0.833 - 44.25 cm/s VD-0.4425 m/ san.. (8)

Taking equation C) for aceleration, . As cace coswt. - - MW (or - a coswt] .(-0.08013) - ( 20 ).... A 2+ 4.59 mlez ... 192 i. The answer is E (option E).