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A 2.22 kg mass on a frictionless surface is connected to a wall by a spring of constant 500 N/m. The mass is pulled 10 cm to the right, then released at t = 0. We will describe the motion according to the formula x t   Acos  t  Please answer each of the following questions. a) What are the values of A, , and ? b) What is the maximum velocity of the block? c) What is the acceleration of the block at t = 0.15 s

User Derrell
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1 Answer

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Answer:

The description for the given question is described in the explanation section below.

Step-by-step explanation:


x = A \ cos(wt + phi)...(equation 1)


dx/dt = v = - Aw \ sin( wt + phi)...(equation 2)

at t = 0,

x = 10 &,

v = 0,

From equation 1, we get


A = 10 \ cos(phi)...(equation 3)


0 = - 10 w \ sin(phi)


phi=0

By using this value in equation 3, we get


A = 10 \ cos(0)


A = 10 \ cm

Now from equation 2, we get


dv/dt = a = -w \ 2A \ cos(wt)...(equation 4)

As we know,


Acceleration = (F)/(m)

at t = 0,

a = -kA/m

= w2 A

w = root(k/m)

= 15 rad/sec

(b)...

From equation 2, we get


Vmax = wA


=1.5 \ m/sec

(c)...

From equation 4, we get


a = - 152* 0.1 cos (15* 0.15)


=-14.15 \ m/sec2

User Mark Cidade
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