A 150-g mass resting on a horizontal frictionless surface is attached to a sprin

Question

A 150-g mass resting on a horizontal frictionless surface is attached to a spring of spring constant 20.0 N/m. It is pulled back a distance of 16.0 cm from its equilbrium position and allowed to oscillate back and forth.

(a) What is the frequency of the oscillations?

(b) What is the speed of the mass each time it passes through the equilibrium position? (Hint: use conservation of energy)

(c) Describe two changes that could be made to the system that would decrease the frequency of the oscillations.

1.

2.

Explanation / Answer

a) w = sqrt(k / m)

2*3.14*f=sqrt(20.0 N/m/0.15)

2*3.14*f=sqrt(133.3)

2*3.14*f =11.54

a)f=1.83Hz

b)in this case, we can use conservation of energy; the PE when the spring is extended equals the KE when the block passes through equilibrium

1/2 k x ^2 = 1/2 m v^2

v = Sqrt[ k x^2/m]

v=sqrt(20*(0.16)^2/0.15)

=sqrt(3.413)

v=1.87m/s

c)Period is inversely proportional to frequency. If period is increased, frequency will decrease

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