Question

A mass of 200 g is tied to a horizontal spring that oscillates with a frequency of 2.0Hz. At time t = 0s, the mass is in the position x = 5.0cm and has a v = -30cm / s. Determine:

a) period

b) angular frequency

c) constant phase amplitude

d) maximum speed

e) maximum acceleration

f) Total mechanical energy

g) position equation for mass

h) position of mass at time t = 0.40s

i) velocity equation for mass

j) at the time the mass passes through

k) the equilibrium position for the first time

l) use the result of part (k) and substitute rn the part (j) to determine the velocity of the mass when it passes through the equilibrium position. Interpret this result.

Answer 1

Solution) m = 200 g

f = 2 Hz

At t = 0 , X = 5 cm and V = - 30 cm/s

(a) Period , T = ?

T = 1/f

T = 1/2

T = 0.5 s

(b) Angular frequency , w = ?

w = 2(pi)f

w = 12.56 rad/s

(c) Constant phase amplitude , A = ?

At t = 0 , X = 5 cm

X = Acos(wt)

5 = Acos(0)

A = 5 cm

(d) Maximum speed , Vmax = ?

Vmax = Aw

Vmax = 5×12.56

Vmax = 62.8 cm/s

(e) Maximum acceleration , amax = ?

amax = A(w^2)

amax = 5×(12.56^2)

amax = 788.76 cm/s^2

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