A 4.40-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 4.40-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 23.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.

A 4.40-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 23.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations (a) Find the force constant of the spring. N/m (b) Find the frequency of the oscillations. Hz (c) Find the maximum speed of the object. m/s (d) Where does this maximum speed occur? (e) Find the maximum acceleration of the object. m/s2 (f) Where does the maximum acceleration occur? (g) Find the total energy of the oscillating system. (h) Find the speed of the object when its position is equal to one-third of the maximum value. m/s mhen s poationis oqual to on-t (i) Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value. m/s2

Explanation / Answer

F = kx

k = F/x = 23/0.200 = 115 N/m

part b )

w = sqrt(k/m)

w = sqrt(115/4.4) = 5.11 rad/s

w = 2pi*f

f = w/2pi

f = 0.814 Hz

part c )

x = Asin(wt)

v = Awcos(wt)

for max coswt = 1

v_max = Aw

v_max = 1.022 m/s

part d )

maximum velocity occurs when it is on lowest point means x = 0

part e )

a = dv/dt

a = -Aw^2sin(wt)

for max = sinwt = -1

amax = Aw^2

amax = 5.227 m/s^2

part f )

it will maximum where velocity is zero , means at amplitude

x = +/- 0.200 m

part h )

Emax = kA^2/2

Emax = 2.3 J

part h )

using energy conservation

kA^2/2 = kx^2/2 + mv^2/2

x = A/3

kA^2(1-1/9) = mv^2

8kA^2/9 = mv^2

v = sqrt(8kA^2/9m)

v = 0.964 m/s

part i )

x = Asin(wt)

x = A/3

1/3 = sin(wt)

wt = 0.3398 rad

t = 0.665

a = -Aw^2(sinwt)

a = -1.65 m//s^2

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