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Two identical masses are attached to two identical springs that hang vertically. The two masses are pulled down and released, but mass B is pulled further down than mass A.

a) Which mass will take a longer time to reach the equilibrium position? Explain.

b) Which mass will have the greater acceleration at the instant of release, or will they have the same acceleration? Explain.

c) Which mass will be going faster as it passes through equilibrium, or will they have the same speed? Explain.

d) Which mass will have the greater acceleration at the equilibrium point, or will they have the same acceleration? Explain.

Respuesta :

Answer:

Lets take mass of A = m

Mass of B = m

Lets take spring constant = K

Displacement of mass A = x

Displacement of mass B= x'

Given that

x ' > x

a)

We know that time period of spring mass system given as

[tex]T= 2\pi\sqrt{\dfrac{m}{K}}[/tex]

Time does not depends on the displacement.

So time will be same for both.

b)

We know that acceleration given as

a= ω² .x

ω=natural frequency  (  ω² = m K)

Here   x ' > x

So Acceleration of mass B is greater than mass A.

c)

Velocity at  equilibrium position

V= ω .x

Here   x ' > x

So velocity of mass B is greater than mass A.

d)

As we know that at equilibrium point acceleration of the mass is zero.So both the mass have same acceleration and the value of this acceleration will be zero m/s².

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