## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 15 - Oscillations - Conceptual Questions: 7

#### Answer

$v_{max} = 28.3~cm/s$

#### Work Step by Step

The total mechanical energy $E$ in the system is constant. At all times during the motion, the sum of the potential energy and the kinetic energy is equal to the total mechanical energy. When all the energy in the system is in the form of kinetic energy, $K = \frac{1}{2}mv_{max}^2$. $\frac{1}{2}mv_{max}^2 = E$ $v_{max} = \sqrt{\frac{2E}{m}}$ Let's suppose the total energy in the system is $2E$. We can find the new value of $v_{max}$ $\frac{1}{2}mv_{max}^2 = 2E$ $v_{max} = \sqrt{\frac{4E}{m}}$ $v_{max} = \sqrt{2}\times \sqrt{\frac{2E}{m}}$ We then find the new value of $v_{max}$: $v_{max} = \sqrt{2}~(20~cm/s)$ $v_{max} = 28.3~cm/s$

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