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FromTheMoon [43]

3 years ago

14

A 140 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless trac

k. The mass is displaced 3 cm from the equilibrium point and released from rest.
(A) Find the period of its motion.
(B) Determine the maximum speed of the block.
(C) What is the maximum acceleration of the block?
(D) Express the position, velocity, and acceleration as functions of time in SI units.

1 answer:

photoshop1234 [79]3 years ago

7 0

Answer:

Explanation:

mass attached m = .14 kg

force constant k = 5N / m

displacement

= amplitude of oscillation

A = .03 m

A ) period of motion = $2\pi\sqrt{\frac{m}{k} }$

= 2 x 3.14 $\sqrt{\frac{.14}{5} }$

T = 1.05 s

B ) maximum speed of block = angular velocity x amplitude

= (2π /T) x A

= (2 x 3.14 x .03) / 1.05

= .1794 m / s

17.94 cm /s

C )

maximum acceleration = angular velocity² x amplitude

= (2π /T)² x A

= (2π /1.05)² x .03

= 1.073 m / s²

D )

position

S = A cos ωt , ω is angular velocity

S = .03 cos(2πt /T)

= .03 cos 5.98 t

v =ω A sin(2πt /T)

= 5.98 x .03 sin5.98t

= .1794 sin5.98t

acceleration = ω²A sin5.98t

= 1.073 sin5.98t

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