A block with a mass of 0.450 kg is connected to a spring, displaced in the posit
ID: 1459123 • Letter: A
Question
A block with a mass of 0.450 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilibrium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 25.0 cm from equilibrium is at t = 0.200 s.
(a) What is the block's period of oscillation?
(b) What is the the value of the spring constant?
(c) What is the block's velocity at t = 0.200 s? (Indicate the direction with the sign of your answer.)
(d) What is the block's acceleration at t = 0.200 s? (Indicate the direction with the sign of your answer.)
PLEASE HELP AND SHOW YOUR WORK!
Explanation / Answer
given mass of the block m = 0.45 kg, block displaced through 50 cm adn released from rest at t = 0 s means
it started from the one of the extream(amplitude) end , and timme to reach 25 cm ( half the amplitude) is 0.2 s
from the data the time period of the spring in horizontal simple harmonic motion is
x= A Cos(wt) = > x= A Cos ((2*pi /T)*t)
25 = 50 Cos ((3.14*0.4)/ T)
T = 1.20 s
a) block's period of oscillaton is T= 1.207 s
b) angular frequency W = 2*pi / T = 2*3.14 / 1.207 = 5.202 rad /s,
W = sqrt(k/m)
= > k= w^2m = 5.202^2*0.45 = 12.17736 N/m
c) the displacement of the oscillator can be written as x= A Cos(wt)
nwo velocity v(t) = - A w Sin (wt) = - 50*5.202 Sin(5.202*0.20)
= -50*5.202 Sin(1.0404)
= - 4.722 m/s
a(t) = - Aw^2 cos (wt ) = - 50*5.202^2 Cos (1.0404) = -1352.81 m/s2
in the negative dierction (-x axis)
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