A particle moves slong the x axis, It is initially at the position 0.160 m, movi

Question

A particle moves slong the x axis, It is initially at the position 0.160 m, moving with velocity 0.100 m/s and acceleration 0.290 m/sSuppose it moves with constant acceleration for 3.50s () Find the position of the particle after this time Find Its velocity at the end of this time interval oscillates in single harmone mot on for a co aroud theequilibrk m position x-O Mint we take the M e partele e d give the same ntial candrons as befeea tratead o having a constant acceleration the following problems are very sensitive to rounding, and you should keep al digits in your calculator Fand the anguiar trequaincy of the oscillation, Hit: in SHM, a is preportional tox Find the amplitude of the oscililstion. Hint: use conservation of energy (e Find its phase constantcasine is used for the equation of motion. Hnt: when taking the innsa of·trig tretion, there are abwarstwo angles but your cale ator all tel you only ong and you must decide which of the two angles you nood rad Find its position after it osciliates for 3.60 s (o) Find its velocity at the end of this 3.60 s time interval

Explanation / Answer

x0 = 0.160 m

v0 = 0.100 m/s

a = - 0.290 m/s^2

t = 3.60 s

(A) x - x0 = v0t + a t^2 / 2

x = 1.36 m

(b) v = v0 + a t = - 0.944 m/s

(c) w = 2 pi / T = 1.75 rad/s

(D) v^2 = w^2 (A^2 - x^2)

0.100^2 = 1.75^2(A^2 - 0.160^2)

A = 0.170 m

(E) x = A sin(wt + phi)

0.160 = 0.170 sin(phi)

ph i= 1.23 rad

(f) x = 0.170 sin[(1.75 x 360) + 1.23]

x = 0.160 m

(g) v = 0.100 m/s

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