A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the s

Question

A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the spring 2.0 cm downward. The mass is then pulled an additional distance of 1.0 cm down and released from rest. Write down its equation of motion. A) y = (0.03 m) sin (22.1 t)
B) y = (0.03 m) cos (22.1 t)
C) y = (0.01 m) sin (22.1 t)
D) y = (0.01 m) cos (22.1 t) A 0.30-kg mass is suspended on a spring. In equilibrium the mass stretches the spring 2.0 cm downward. The mass is then pulled an additional distance of 1.0 cm down and released from rest. Write down its equation of motion.

Explanation / Answer

The general equation of simple harmonic motion is

y = a cos (? t )..... or y = a sin (? t )....

but here y = a cos (? t )...because at t=0 amplitude is maximum

Here amplitude 'a '= 0.01 m
m = 0.3 kg
x = 0.02 m
Force constant k is given by,
k = mg / x = 0.3 x 9.8/ 0.02 = 147 N/m

The time period of oscillation is given by ,
T = 2 ? ?( m / k )
? = 2 ? / T = ? ( 147 / 0.3 )
? = 22.1

then y = (0.01m) cos (22.1 t )......ans is (D)

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