Calculating Frequency of Oscillation for Different Masses on a Spring

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SUMMARY

The frequency of oscillation for a mass-spring system is inversely related to the square root of the mass. A 0.25 kg mass oscillates at 1.0 Hz, while a 0.50 kg mass results in a frequency of 0.71 Hz. The error in the initial calculation stemmed from incorrectly squaring the mass ratio instead of taking the square root. The correct formula is f2 = f1 * √(m1/m2), leading to the accurate frequency for the 0.50 kg mass.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Basic knowledge of oscillatory motion and frequency calculations
  • Familiarity with the formula for frequency in mass-spring systems
  • Ability to manipulate square roots and ratios in mathematical equations
NEXT STEPS
  • Study the derivation of the frequency formula for mass-spring systems
  • Explore the effects of varying spring constants on oscillation frequency
  • Learn about damping effects in oscillatory systems
  • Investigate real-world applications of mass-spring oscillators in engineering
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Physics students, mechanical engineers, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

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"A 0.25kg mass on a vertical spring oscillates with a frequency of 1.0 Hz. Suppose a 0.50kg mass were instead suspended from the same spring. What would be the frequency of oscillation in this case?"

This is what I did:
[tex]\frac{f_{1}}{f_{2}} = \frac {\sqrt{\frac{k}{m_{1}}}}{\sqrt{\frac{k}{m_{2}}}}[/tex]
[tex]\frac{f_{1}}{f_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}[/tex]
[tex]f_{2} = \frac{1}{4} f_{1}[/tex]
[tex]f_{2} = 0.25 Hz[/tex]
however, the answer is 0.71 Hz. What did I do wrong?
 
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You got the ratio [itex]m_2/m_1[/itex] wrong. The ratio is 2, which means that [itex]\sqrt{m_2/m_1}=\sqrt{2}[/itex], but you have 4.

So in short I think you squared the 2 when you should have taken the square root.
 
You are absolutely correct! I didn't realize I did that. Thanks for the prompt reply!
 

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