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Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance 6.7 k/m as the two systems reach equilibrium. Then both springs are compressed and released. Which one oscillates faster?

a.) The spring with the heavier ball,
b.) Springs oscillate with the same frequency,
c.) The spring with the light ball.

Answer :

Answer:

option (b)

Explanation:

Let both teh springs are extended by y.

For first spring

m1 x g = k1 x y

where, m1 is the mass of first ball and k1 be the spring constant of first spring

[tex]y = \frac{m_{1}g}{k_{1}}[/tex]    .... (1)

For second spring

m2 x g = k2 x y

where, m2 is the mass of second ball and k2 be the spring constant of second spring

[tex]y = \frac{m_{2}g}{k_{2}}[/tex]    .... (2)

The frequency of the body is given by

[tex]f = \frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

So, from equation (1) and equation (2), the ratio of k is to m is same, so the frequency of both the balls is same.

b.) Springs oscillate with the same frequency

Let both the springs are extended by y.

For first spring

[tex]m_1 * g = k_1 * y[/tex]

where, m₁ is the mass of first ball and k₁ be the spring constant of first spring

[tex]y=\frac{m_1*g}{k_1}[/tex]    .......... (i)

For second spring

[tex]m_2 *g = k_2 *y[/tex]

where, m₂ is the mass of second ball and k₂ be the spring constant of second spring

[tex]y=\frac{m_2*g}{k_2}[/tex]    ........... (ii)

The frequency of the body is given by

[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m} }[/tex]

So, from equation (i) and equation (ii), the ratio of k is to m is same, so the frequency of both the balls is same.

Therefore, option b is correct.

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