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A mass m is suspended from a spring. Its...

A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into two halves and the same mass is suspended from one of the pieces of the spring. The frequency of oscillation of the mass will be

A

`sqrt(2) f`

B

`(f)/(2)`

C

f

D

2f

Text Solution

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The correct Answer is:
To solve the problem step by step, we will use the formula for the frequency of oscillation of a mass-spring system and analyze how cutting the spring affects the spring constant. ### Step-by-Step Solution: 1. **Understand the Frequency Formula**: The frequency of oscillation (f) of a mass (m) attached to a spring with spring constant (k) is given by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] 2. **Initial Frequency**: Let the initial frequency of oscillation when the mass is suspended from the original spring (with spring constant k) be: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] 3. **Effect of Cutting the Spring**: When the spring is cut into two halves, each half will have a new spring constant. The spring constant of a spring is inversely proportional to its length. When the length of the spring is halved, the spring constant doubles. Therefore, the spring constant of one half (let's call it \( k' \)) becomes: \[ k' = 2k \] 4. **New Frequency with Half Spring**: Now, if the same mass (m) is suspended from one half of the spring, the new frequency \( f' \) can be calculated using the new spring constant \( k' \): \[ f' = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}} \] 5. **Relate New Frequency to Old Frequency**: We can relate the new frequency \( f' \) to the original frequency \( f \): \[ f' = \frac{1}{2\pi} \sqrt{\frac{2k}{m}} = \sqrt{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \sqrt{2} \cdot f \] 6. **Final Result**: Therefore, the frequency of oscillation when the mass is suspended from one half of the spring is: \[ f' = \sqrt{2} \cdot f \] ### Conclusion: The frequency of oscillation of the mass when suspended from one piece of the cut spring is \( \sqrt{2} \) times the original frequency \( f \).

To solve the problem step by step, we will use the formula for the frequency of oscillation of a mass-spring system and analyze how cutting the spring affects the spring constant. ### Step-by-Step Solution: 1. **Understand the Frequency Formula**: The frequency of oscillation (f) of a mass (m) attached to a spring with spring constant (k) is given by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} ...
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