If x = .4sin(5t + .2) and K1(200 N/m) and K2(160 N/m) are two springs placed end

Question

If x = .4sin(5t + .2) and K1(200 N/m) and K2(160 N/m) are two springs placed end to end.
a) What is the ktotal of both springs? What is the mass of the block?
b) What is the xi, vi, ai ?
c) What is the absolute value of the maximum values of x, v, a, K.E., U?
d) What is the period T, angular frequency ?? and frequency f of the block?
e) How many seconds will it take the block to 1st reach the most compressed position?
f) How many seconds will it take for the KE of block to equal twice PE of spring?

Explanation / Answer

x = .4sin(5t + .2)

v =dx/dt = 2*cos(5t+.2)

a = dv/dt = -10*sin(5t+0.2)

a.) ktotal = K1*K2/(K1+K2) = 88.89 N/m

   = 5

=> m/k = 5 => m = 25k = 2222.22 kg

b.) xi = 0.4*sin(0.2) = 0.08 m

vi = 0.4**cos(0.2) = 1.96 m/s

ai = -10*sin(0.2) = -1.99 m/s2 -2 m/s2

c.) xmax = 0.4 m

vmax = A = 2 m/s

amax = A2 = 10 m/s2

K.E.max = (1/2)mvmax2 = 0.5*2222.22*22 = 4444.44 J = 4.44 kJ

Umax = K.E.max = 4.44 kJ

d.) angular frequency = = 5 rad/s

T = 2/ = 1.26 s

f = /2 = 0.796 Hz

e.) Most compressed point i.e. xmax

=> sin(5t+0.2) = 1 ( for lowest t

=> 5t+0.2 = /2 => t = 0.274 s

f.) We know that

U + K.E. = 4.44 kJ

2U =K.E.

3U = 4.44 kJ

=> U = 4.44/3 kJ

=> K.E. = 2U

K.E. = 4444.44*2/3 = (1/2)mv2

=> v2 = 2*2*2/3 => v = 1.633 m/s

1.633 = 2*cos(5t+0.2)

=> 0.615 = 5t+0.2 => t = 0.083 s

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