13.5 A spring with a spring constant of 450 N/m is stretched 20 cm from the equi

Question

13.5 A spring with a spring constant of 450 N/m is stretched 20 cm from the equilibrium position. a) What is the magnitude of the spring force at x = 20 cm? b) If a 5 kg mass is attached to the spring, what will the maximum acceleration be if the spring is released from the x = 20 cm stretched position? c) Will the acceleration be the same as the spring passes the x = 10 cm position? If not, what is the acceleration at that instant? d) What is the frequency of oscillation of the spring?

13.6 A 12 kg mass is attached to a spring with a spring constant of 1600 N/m. The spring is stretched 7 cm from equilibrium and released at t = 0 s. a) Write the equation for the position as a function of time. b) What is the position of the spring at t = 2.5 s?

Explanation / Answer

13.6 :

A = amplitude = 7 cm = 0.07 m

k = spring constant = 1600 N/m

m = mass attached = 12 kg

w = angular frequency = sqrt(k/m) = sqrt(1600/12) = 11.6 rad/s

equation is given as

x = A Coswt

x = (0.07) Cos(11.6t)

b)

at t = 2.5 s

x = (0.07) Cos(11.6t)

x = (0.07) Cos(11.6 x 2.5)

x = - 0.052 m

13.5 )

a)

k = spring constant = 450 N/m

x = stretch in spring = 20 cm = 0.20 m

magnitude of spring force is given as

Fs = kx

Fs = 450 x 0.20

Fs = 90 N

b)

m = mass attached = 5 kg

k = spring force = 450

w = angular frequency = sqrt(k/m) = sqrt(450/5) = 9.5 rad/s

maximum acceleration is given as

a = Aw2 = (0.20) (9.5)2 = 18.05 m/s2

c)

acceleration will not be same , since acceleration depends on the stretch in the spring.

a = w2 x = (0.10) (9.5)2 = 9.025 m/s2

d)

w = angular frequency = sqrt(k/m) = sqrt(450/5) = 9.5 rad/s

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