A 1.5kg object oscillates at the end of a vertically hanging light spring once e

Question

A 1.5kg object oscillates at the end of a vertically hanging light spring once every 0.40s .

A. Write down the equation giving its position y (+ upward) as a function of time t. Assume the object started by being compressed 19cm from the equilibrium position (where y = 0), and released.

a. y(t)=(0.19m)*cos(t/0.40s)   

b. y(t)=(0.19m)*cos((2*pi*t)/0.40s)   

c. y(t)=(0.19m)*cos(0.40s*t)      

d. y(t)=(0.19m)*sin((2*pi*t)/0.40s)

B. How long will it take to get to the equilibrium position for the first time?

C. What will be its maximum speed?

D. What will be the object's maximum acceleration?

E. Where will the object's maximum acceleration first be attained?

- release point

- equilibrium point

Explanation / Answer

A 1.60-kg object oscillates from a vertically hanging light springonce every 0.55s.
(a) Write down the equation giving its position y (taking upward asthe +y) as a function of time t, assuming it started by beingcompressed 16 cm from the equilibrium position (where y = 0), andreleased.
(b) How long will it take to get to the equilibrium position for the first time?
(c) What will be its maximum speed?
(d) What will be its maximum acceleration, and...
(e) where will it first be attained?

Answers

A)

we have A = 16 cm = 0.16m, T = 0.55 s

y = A cos (2pi t/T) = 0.16 m cos ( 2 pi t / 0.55s ) = 0.16 m cos( 11.4 t)

b.

t = 1/4 T = 1/4 (0.55s) = 0.14 s

c.

v max = w A = 2 pi / T (A) = ( 2 pi / 0.55s ) (0.16m) = 1.8m/s

d.

a max = (w^2) A = 20.89 (m/s^2)

e.

*same answer as part d*

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