# A mass m at the end of a spring vibrates with a frequency of 0.70 Hz . When an additional 660 g...

## Question:

A mass {eq}m {/eq} at the end of a spring vibrates with a frequency of {eq}0.70 \ Hz {/eq}. When an additional {eq}660 \ g {/eq} mass is added to {eq}m {/eq}, the frequency is {eq}0.60 \ Hz {/eq}. What is the value of {eq}m {/eq}?

## Spring-mass System:

The frequency of vibration of the spring-mass system depends on two parameters: one is the modulus of the spring and the second is the mass attached at the moving end of the spring. For the smaller mass and the high modulus of the spring, the coiled spring will vibrate with the greater frequency. For the given spring, the modulus of the spring remains constant, therefore, the frequency of vibration changes due to the change in the suspended mass.

## Answer and Explanation: 1

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Given data:

• {eq}f=\rm 0.70 \ Hz {/eq} is the frequency of vibration with mass m
• {eq}f'=\rm 0.60 \ Hz {/eq} is the frequency of vibration after...

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#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.