Question

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is ? = 2 /3 cos(8?)− 1 /6 sin(8?), ? ≥ 0, where ? is measured in centimeters and ? in seconds. (Take the positive direction to be downward.)

g. How far from its equilibrium position does the mass travel?

h. Find the total distance traveled by the mass during the first 4? seconds.

i. When is the mass speeding up when heading downward?

Answer 1

Solution-

Initially at the bottom-most point,

(g)

Find position at t=0, at the extreme point

2/3 cm from the mean position.

It is the maximum extension of the band.

(h)

The time taken to come from bottom extreme to mean position at the first run is the fourth part of the time period,

Calculate the time to reach a mean position,

s=0

It is the fourth part of the time period,

T = 4 x 0.1657= 0.6628 s

Total periods in given time of 4X3.14,

The total distance will be,

Dt=19XDT

Distance in one time period is DT= 4X2/3 = 2.67 cm

Dt=19X2.67=50.7 cm

Therefore, the total distance in 4pi seconds is 50.7 cm.

(h)

For the first time at the topmost point where acceleration is towards downward,

while the object moves down it speeds up,

So for the first time after starting from a bottom-most point,

Half of the time period = 0.5 T = 0.5X0.6628 = 0.3314 s

Note: curve for position s(t)

For an initial passing time,

v-t curve

a-t curve