A block on a frictionless table is connected as shown in $FIGUREP15.74$to two springs having spring constants $k_1$and $k_2$. Show that the block's oscillation frequency is given by

$f=\sqrt{f_1^2+f_2^2}$

where $f_1$and $f_2$are the frequencies at which it would oscillate if attached to spring $1$or spring $2$alone.

FIGURE P15.74

Hooke's Law states that every thing that causes a spring to extend or compress puts an elastic force on that object. The$x$-component of the force the spring exerts on the item if the object strains the spring in the $x$-direction is:

$F_{\mathrm{S}\text{on}\mathrm{O},x}=-kx$

where$k$is the spring constant in newtons per metre, which is a measure of the spring's (or any elastic object's) stiffness, and$x$is the stretch/compression distance (not the total length of the object). The spring's elastic force on the object is in the opposite direction of how it was stretched (or compressed), hence the negative sing in front of $kx$. The spring is then acted upon by the object:

$F_{oons,x}=+kx$

In simple harmonic motion, the frequency of an oscillator is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

It's worth noting that $f$ is solely affected by the mass $m$ and the force constant $k$, rather than the amplitude.

$k_1$and $k_2$are the spring constants for the two springs.

$Figure15.74$depicts a mass $m$block on a frictionless table connected to two springs.

If the block is only attached to springs$1$or $2$, the oscillation frequencies are $f_1$and $f_2$, respectively.

We are asked to show that the block's oscillation frequency is given by$f=\sqrt{f_1}+f_2$.

The block-double spring system is depicted in the diagram below when the block is out of balance. Both springs want to return to their equilibrium configuration, therefore the left spring will push the block to the right and the right spring will pull the block to the left, resulting in both forces acting to the right on the block. The block's net force is then calculated.

$F_{net}=F_1+F_2$

where the magnitudes of the forces exerted by both springs are obtained in equation $(1)$and $f_1$and $f_2$are the magnitudes of the forces exerted by both springs

$F_{\text{net}}=k_1\Delta x+k_2\Delta x$

$=\left(k_1+k_2\right)\Delta x$

localid="1650300945352" $=k_{eff}∆x$

Where localid="1650267212984" $k_{eff}=k_1{k}_2$is the effective spring constant of the system.

The frequency of oscillation of the block if it is attached to spring $1$alone is found from Equation$\left(2\right)$

$f_1=\frac{1}{2\pi }\sqrt{\frac{k_1}{m}}$

Rearrange and solve for $k_1$:

$f_1^2=\frac{1}{4{\pi }^2}\frac{k_1}{m}$

$k_1=4{\pi }^{2}m{f}_1^2$

Similarly, the frequency of oscillation of the block if it is connected to spring $2$alone is:

Rearrange and solve for $k_2$

${f_2}^2=\frac{1}{4{\pi }^2}\frac{k_2}{m}$

$k_2=4{\pi }^{2}m{f}_2^2$

Equation$\left(2\right)$can also be used to calculate the frequency of oscillation of the block and two springs in terms of the effective spring constant of the system:

$f=\frac{1}{2\pi }\sqrt{\frac{k_{\mathrm{eff}}}{m}}$

$=\frac{1}{2\pi }\sqrt{\frac{k_1+k_2}{m}}$

Substitute for $k_1$and $k_2$from equation $\left(3\right)$and $\left(4\right)$

$f=\frac{1}{2\pi }\sqrt{\frac{4{\pi }^{2}m{f}_1^2+4{\pi }^{2}m{f}_2^2}{m}}$

$=\frac{1}{2\pi }\sqrt{\frac{4{\pi }^2\cancel{m}\left(f_1^2+f_2^2\right)}{\cancel{m}}}$

$=\frac{1}{2\pi }\sqrt{4{\pi }^2\left(f_1^2+f_2^2\right)}$

$=\sqrt{f_1^2+f_2^2}$