# effect of mass of spring on period of oscillation

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 ($T$)?

• 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 – Red Floyd Feb 18 '17 at 1:44
• Yes,the period will be different, you should watch the Walter lewin lectures on waves and vibration. probably lecture no 2 or 3 – Paul Feb 18 '17 at 1:55
• Possible duplicates: physics.stackexchange.com/q/64934/2451 , physics.stackexchange.com/q/78711/2451 , and links therein. – Qmechanic Feb 18 '17 at 8:40

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.