Answer:
[tex]A'=2A[/tex]
Explanation:
According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:
[tex]E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}[/tex]
When the spring is in its equilibrium position, that is [tex]x=0[/tex], the object speed its maximum. So, we have:
[tex]\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}[/tex]
In order to double its maximum speed, that is [tex]v'{max}=2v_{max}[/tex]. We have:
[tex]A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A[/tex]
The factor we must increase the amplitude of the vibration to double its maximum speed is 2.
Apply the principle of conservation of energy to determine the amplitude;
[tex]U = K.E\\\\\frac{1}{2} KA^2 = \frac{1}{2}mv^2\\\\KA^2 = mv^2\\\\A^2 = \frac{m}{K} v^2\\\\\frac{A_1^2}{v_1^2 } = \frac{A_2^2}{v_2^2 } \\\\A_2 = \frac{v_2 A_1}{v_1} \\\\A_2 = \frac{2v_1 \times A_1}{v_1} \\\\A_2 =2A_1[/tex]
Thus, the factor we must increase the amplitude of the vibration to double its maximum speed is 2.
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