A particle moves along the x axis. It is initially at the position 0.300 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.300 m, moving with velocity 0.070 m/s and acceleration -0.260 m/s^2. Suppose it moves with constant acceleration for 3.40 (a) Find the position of the particle after this time. (b) Find its velocity at the end of this time interval. m/s We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 3.40 s around the equilibrium position x = 0. (c) Find the angular frequency of the oscillation. (d) Find the amplitude of the oscillation. (e) Find its phase constant phi_0 if cosine is used for the equation of motion. (f) Find its position after it oscillates for 3.40 s. (g) Find its velocity at the end of this 3.40 s time interval. m/s

Explanation / Answer

a)

from

X=Xo+Vot+(1/2)at2=0.3+0.07*3.4+(1/2)(-0.26)*3.42=-0.965 m

b)

from V=Vo+at=0.07+(-0.26)*3.4

V=-0.814 m/s

c)

acceleration of the particle

a=-W2x

-0.26=-W2*0.3

W=0.93 rad/s

d)

By conservation of energy

(1/2)KA2=(1/2)mV2+(1/2)Kx2

since K=W2m

(1/2)mW2A2=(1/2)mV2+(1/2)mW2x2

0.932*A2=0.072+0.932*0.32

A=0.3093 m

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