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Consider a massless spring system which is hanging vertically. Attach a mass $M$ and set it into simple harmonic motion. Let the period with which the mass oscillates be $T$.

We assume that the spring is massless in most cases. Would taking effect of the non-zero mass of the spring affect the time period ($T$)?

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  • $\begingroup$ As far as i remember, ke of spring with mass comes out to be approximately 1/3mv² (or 1/6) . Now write total energy = ke(spring)+ke(block)+pe(block). Differentiate. You will get the new time period $\endgroup$ Feb 18, 2017 at 1:44
  • $\begingroup$ Yes,the period will be different, you should watch the Walter lewin lectures on waves and vibration. probably lecture no 2 or 3 $\endgroup$
    – Paul
    Feb 18, 2017 at 1:55
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/64934/2451 , physics.stackexchange.com/q/78711/2451 , and links therein. $\endgroup$
    – Qmechanic
    Feb 18, 2017 at 8:40

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The effective mass of the spring when oscillating alone is $m^*=\frac13m$ where $m$ is its actual mass. You would add $m^*$ to the mass $M$ of the object hanging from it in order to calculate the period $T$ of oscillation. $T \propto \sqrt{M}$ so the mass of the spring increases the period of oscillation.

See wikipedia article Effective Mass of Spring in Mass-Spring System. The reference there (A Measurement of the Effective Mass of Coil Springs) states that this theoretical value of $m^*$ for an unloaded spring $(M/m = 0)$ holds quite well for values of $M/m < 7$ but above that limit decreases and becomes -ve.

See also Effective mass in Spring-with-mass/mass system and What will be different if the spring is not massless?

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    $\begingroup$ The factor of 1/3 is an approximation, assuming the spring mass is small compared with the attached mass. In general, the system will not oscillate with a single period T, because it now has (in theory at least) an infinite number of vibration modes where different parts of the spring are vibrating with different amplitudes, and the motion will be some linear combination of all of those modes. $\endgroup$
    – alephzero
    Feb 18, 2017 at 5:21
  • $\begingroup$ I think one would naturally ask, is it still SHM? $\endgroup$ Feb 18, 2017 at 5:23
  • $\begingroup$ @C.TowneSpringer The formula does not change, only the value of what you substitute as mass changes. Therefore, the system is still shows SHM if you allow some amount of tolerance. $\endgroup$
    – Yashas
    Feb 18, 2017 at 5:48

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