A mass m = 4.6 kg is attached to a vertical spring with k = 192 N/m and is set i

Question

A mass m = 4.6 kg is attached to a vertical spring with k = 192 N/m and is set into motion.

(a) What is the frequency of the oscillation?
Hz

(b) If the amplitude of the oscillation is 4.0 cm, what is the maximum value of the velocity?
m/s

(c) How long does it take the mass to move from y = 1.5 cm to y = 2.5 cm?
s

(d) If the mass is oscillating with a maximum speed of 45 m/s, what is the amplitude?
m

(e) If the spring constant is increased by a factor of two and the maximum kinetic energy of the mass is the same, by what factor does the amplitude change?
% of the original amplitude

Explanation / Answer

a)

frequency, f = (1/(2*pi))*sqrt(k/m)

= (1/(2*pi))*sqrt(192/4.6)

= 1.03 Hz <----answer

b)

0.5*k*x^2 = 0.5*m*Vmax^2 <------ using conservation of energy

So, 192*0.04^2 = 4.6*Vmax^2

So, Vmax = 0.258 m/s

c)

y = A*cos(2*pi*f*t)

So, 1.5 = 4*cos(2*pi*1.03*t1)

So, t1 = 0.183 s

For 2.5 = 4*cos(2*pi*1.03*t2)

So, t2 = 0.138 s

So, time taken = t1 - t2 = 0.183 - 0.138 = 0.045 s

d)

Using equation of motion,

0.5*k*A^2 = 0.5*m*Vmax^2

So, 192*A^2 = 4.6*45^2

So, A = 6.97 m

e)

0.5*k*A^2 = KE

KE remains same.

So, as k is increased by a factor of 2, A must be decreased by a factor of 1/sqrt(2) = 0.707 times

= 70.7 percent of the original

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