Question

A 0.580-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 13.0 cm. (Assume the position of the object is at the origin at

t = 0.)

(a) Calculate the maximum value of its speed.
cm/s

(b) Calculate the maximum value of its acceleration.
cm/s²

(c) Calculate the value of its speed when the object is 11.00 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 11.00 cm from the equilibrium position.
cm/s²

(e) Calculate the time interval required for the object to move from x = 0 to x = 5.00 cm.
s

Answer 1

1.Sol::

Given:

Mass (m) = 0.580 kg
spring constant (k) =8 N /m
amplitude = 13.0 cm

Let us take x(t) = Asin(ωt - φ)
Where φ = 0 here

Then velocity v(t) = Aωcos(ωt)
and acceleration a(t) = -Aω²sin(ωt)

And ω = √(k/m)
= √(8 / 0.58)
= 3.714 rad/s

a)
Now
vmax = Aω
= 13 cm * 3.714 rad/s
= 48.28 cm/s

b)
amax = Aω²
= 13 cm * (3.714 rad/s)²
= 179.32 cm/s²

c)
Given x(t)= 11 cm

=> x(t) = 11 cm = 13 cm * sin(3.714t)
=> 3.714t = arcsin(11/13)
= 1.008 rads
t = 1.008/3.714
= 0.271 s

Then
v(0.271) = 13 * 3.714* cos(3.714 * 0.271)
= 25.8 cm/s

d)
Now
a(0.271) = -13 * (3.714)² * sin(3.714 * 0.271)
= ±151.52 cm/s²
depending on whether it is coming or going.

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