A 220 g block is attached to a horizontal spring and executes simple harmonic mo

Question

A 220 g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.150 s. If the total energy of the system is 4.00 J. (a) Find the force constant of the spring N/m (b) Find the amplitude of the motion. m A block-spring system oscillates with an amplitude of 3.65 cm. The spring constant is 250 N/m and the mass of the block is 0.500 kg. (a) Determine the mechanical energy of the system. Your response is within 10% of the correct value. This may be due to round off error, or you could have a mistake in your calculation. Carry out all minimize round off error. J (b) Determine the maximum speed of the block. m/s (c) Determine the maximum acceleration. m/s^2

Explanation / Answer

(a)The angular frequency of the spring is

w = (2/T)

T = 0.150 s

or w = (2 * 3.14/0.150) = 41.87 rad/s

we know that

w = (k/m)1/2

or w2 = (k/m)

or k = w2 * m

m = 220 g = 220 * 10-3 kg

or k = (41.87)2 * 220 * 10-3 = 385.68N/m

(b)The total energy of the system is

E = (1/2)kA2

or A = (2E/k)1/2

E = 4.0 J

or A = (2 * 4.0/385.68)1/2 = 0.144 m

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