Question

A metal cylinder with a mass of 1.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The cylinder is pulled to a distance of 0.200 mfrom its equilibrium position, held in place with a force of 17.0 N, and then released from rest. It then oscillates in simple harmonic motion. (The cylinder oscillates along the x-axis, where x = 0 is the equilibrium position.)

(a) What is the spring constant (in N/m)?

_____ N/m

(b) What is the frequency of the oscillations (in Hz)?

_____ Hz

(c) What is the maximum speed of the cylinder (in m/s)?

_____ m/s

(d) At what position(s) (in m) on the x-axis does the maximum speed occur?

x = ± ______ m

(e) What is the maximum acceleration of the cylinder? (Enter the magnitude in m/s².)

______ m/s²

(f) At what position(s) (in m) on the x-axis does the maximum acceleration occur?

x = ± ______ m

(g) What is the total mechanical energy of the oscillating spring–cylinder system (in J)?

______ J

(h) What is the speed of the cylinder (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium?

______ m/s

(i) What is the magnitude of the acceleration of the cylinder (in m/s²) when its position is equal to one-third of the maximum displacement from equilibrium?

_______ m/s²

Answer 1

a)

F = force applied to hold the cylinder in place = 17 N

x = stretch of the spring = 0.200 m

k = spring constant = ?

Using the equation for force of spring

F = k x

17 = (0.200) k

k = 85 N/m

b)

m = mass of the cylinder attached = 1.20 kg

frequency of oscillation is given as

f = (1/(2)) sqrt(k/m)

f = (1/(2(3.14))) sqrt(85/1.20)

f = 1.34 Hz

c)

v_max = maximum speed

A = amplitude = 0.200 m

angular frequency is given as

w = 2f = 2 (3.14) (1.34) = 8.42

maximum speed is given as

v_max = A w

v_max = (0.200) (8.42)

v_max = 1.7 m/s

d)

maximum speed occurs at the equilibrium position.

hence x = 0

e)

maximum acceleration is given as

a_max = A w²

a_max = (0.200) (8.42)²

a_max = 14.2 m/s²