Question

By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum speed during a vibration? A is the old amplitude and A′ is the new one.

Answer 1

Answer:

$A'=2A$

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:

$E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}$

When the spring is in its equilibrium position, that is $x=0$, the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

In order to double its maximum speed, that is $v'{max}=2v_{max}$. We have:

$A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A$