Question

The cart is then pushed up the ramp, compressing the spring a distance A from equilibrium. When

Answer 1

When the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is $\sqrt{\frac{1}{2} } \ T_1$.

Period of the mass oscillation

The period of the oscillation of the mass or cart on the spring is given by the following formula;

$T = 2\pi \sqrt{\frac{m}{k} }$

at a costant mass;

$T_1\sqrt{k_1} = T_2\sqrt{k_2} \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{k_2} }$

when spring constant is doubled, k2 = 2k1. the new period, T2 is determined as follows;

$T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2k_1} } \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2} \times \sqrt{k_1} } \\\\T_2 = \frac{T_1}{\sqrt{2} }\\\\T_2 = \sqrt{\frac{1}{2} } \ T_1$

Thus, when the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is $\sqrt{\frac{1}{2} } \ T_1$.

Learn more about period of oscillation here: brainly.com/question/20070798

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