A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.6 cm. (Assume the position of the object is at the origin at

t = 0.)

(a) Calculate the maximum value of its speed.

(b) Calculate the maximum value of its acceleration.

(c) Calculate the value of its speed when the object is 8.60 cm from the equilibrium position.

(d) Calculate the value of its acceleration when the object is 8.60 cm from the equilibrium position.

(e) Calculate the time interval required for the object to move from x = 0 to x = 4.60 cm.

Can I please get the answers in steps.

Explanation / Answer

here,

mass of the object , m = 0.4 kg

spring constant , K = 8 N/m

amplitude , A = 10.6 cm

A = 0.106 m

teh angular frequency , w = sqrt(K/m)

w = sqrt (8/0.4)

w = 4.47 rad/s

(a)

the maximum value of its speed , Vmax = A * w

Vmax = 0.106 * 4.47

Vmax = 0.47 m/s

the maximum value of its speed is 0.47 m/s

(b)

the maximum value of its acceleration , amax = A * w^2

amax = 0.106 * 4.47^2

amax = 2.12 m/s^2

the maximum value of its acceleration is 2.12 m/s^2

(c)

let the value of its speed when the object is 8.60 cm from the equilibrium position be v
distance x = 8.6 cm

x = 0.086 m

using conservation of energy ,

0.5 * m * v^2 = 0.5 * k * x^2

0.4 * v^2 = 8 * 0.086^2

v = 0.38 m/s

the value of its speed when the object is 8.60 cm from the equilibrium position is 0.38 m/s

(d)

let the value of its acceleration when the object is 8.60 cm from the equilibrium position be a

a = x * w^2

a = 0.086 * 4.47^2

a = 1.72 m/s^2

the value of its acceleration when the object is 8.60 cm from the equilibrium position is 1.72 m/s^2

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