You are watching an object with mass m that is attached to a spring with force c

Question

You are watching an object with mass m that is attached to a spring with force constant k that is moving horizontally with simple harmonic motion. When the object is displaced 0.500 m to the right of its equilibrium position, it has a velocity of 2.00 m/s to the right, and an acceleration of 8.00 m/s 2 to the left.

A)  Recalling Hooke’s law for the force exerted by the spring on the mass, and Newton’s second law relating this force to the mass and acceleration, calculate the ratio m / k

B) Calculate the period of oscillation for this mass and spring.

C) Using conservation of energy, calculate the amplitude of oscillation A .

Explanation / Answer

A)

velocity at position x is

v = w*sqrt(A^2-x^2)

2 = w*sqrt(A^2-0.5^2)

and also accelaration a = -w^2*x

8 = -w^2*0.5

w^2 = -16 rad/sec

w^2 = k/m = 16 rad/s

m/k = 1/16

B) period of oscillation is T = 2*pi*sqrt(m/k) = 2*3.142*sqrt(1/16) = 1.57 sec

C) using conservation of energy

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

dividing with k on both sides

0.5*(m/k)*v^2 + 0.5*x^2 = 0.5*A^2

(1/16)*2^2 + 0.5^2 = A^2

Amplitude is A = 0.707 m

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