Question

Consider 100 g mass hung from a spring at equilibrium (y = 0). Write the equation of oscillation, and sketch of the mass’ position y(t) for the following situations. Be sure to correctly identify each function drawn.

a) Suppose the mass is displaced upward from the equilibrium position by 10 cm and then released from this position at time t = 0. A period of oscillation of 2 second was observed. Draw this on the first graph.

(b) Suppose the mass is displaced downward by 5 cm and then released from this position at time t = 0. Draw this on the same graph. (

c) Suppose the mass is decreased to 25 g and is allowed to come to its new equilibrium position (which we again call y = 0). It is then displaced upward from the new equilibrium position by 10 cm and then released from this position at time t = 0. Draw this on the second graph. (

d) Suppose you started observing the same oscillation but 0.25 seconds after it was released. Draw this on the same graph. Indicate on the graph the time delay between this oscillation and the previous one.

Answer 1

General equation of oscillations is

Asin(wt) if starting from mean position
Acos(wt) if starting from extreme position

A is amplitude
w = 2/T
T is periodic time

a)

As 10cm is the initial displacement, this will become the amplitude. Periodic time is given as 2 seconds.

It will look like this

w = 2/2 =

The equation will be 10cos(t)

b)

The amplitude becomes -5cm but periodic mass remains same, because spring and mass are still the same.

The equation will be -5cos(t)

c)

k/m = w^2

k/m = (2/T)^2

so, m T^2

In this case, mass is decreased to 25g

m1/m2 = (T1/T2)^2

100/25 = (2/T2)^2 = 4/T2^2

100/100 = T2^2

T2 = 1 sec

w = 2/T = 2/1 = 2

so, the graph will look like this

The equation will be 10cos(2t)

d)

As it is the same oscillation, but starting only 0.25 sec later

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