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/s^{2}

(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/s^{2}

(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.

Hope this answer helps you

Please RATE my answer

Thank you

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