A mass, m = 0.50 kg, is attached to a horizontal spring of spring constant, k =

Question

A mass, m = 0.50 kg, is attached to a horizontal spring of spring constant, k = 200 N/m. The mass is given an initial displacement of 0.015 m and an initial velocity of +0.40 m/s. a) Find the period, amplitude, and phase angle of the motion. b) Write equations for the displacement, velocity and accelerations as a function of time. A mass oscillating in simple harmonic motion starts at x = A and has a period of T. At what time, as a fraction of T, does the object first pass through x = (1/2)A? An object on a spring oscillates with a period of 0.80 s and an amplitude of 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of motion at t = 2.0 s? An 83 kg student hangs from a bungee cord with a spring constant of 270 N/m. The student is pulled down to a point where the cord is 5.0 m longer than its unstretched length, then released. Where is the student, and what is his velocity 2.0 s later? Question4

Explanation / Answer

(A) T= 2 pi sqrt( m / k )

= 2pi sqrt(0.50 / 200) = 0.314 sec .....Time period

Total energy = k x^2 /2 + m v^2 /2 = k A^2 /2

(200 x 0.015^2) + (0.50 x 0.40^2) = (200) A^2

A = 0.025 m ........Amplitude

x = A sin(wt + phi)

0.015 = 0.025 sin(phi)

phi = 0.64 rad ......Ans(phase angle)

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x = A cos(wt)

A/2 = A cos(2 pi t / T)

2 pi t / T = pi/ 3

t = T / 6

--------------------

x = A sin(wt + phi)

- 5 = 10 sin(phi)

phi = - 3.67 rad

x = 10 cm sin(2pi x 2 /0.80 - 3.67)

x = 2.65 cm

direction -> to the right

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for equilibrium position,

kx - m g = 0

x0 = (83 x 9.81) / 270 = 3.02 m

A = 5 - 3.02 = 1.98 m

w = sqrt(k/m) = 1.80 rad/s

x = A cos(wt) = 1.98 cos(1.80 t)

for t = 2s ,

x = -1.78 m

so student is 5 + 1.78 = 6.78 m above the lowest point. .....Ans

v = - 3.564 sin(1.80t)

t = 2s .

v = 1.58 m/s
hence velocity is 1.58 m/s downwrad.

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