Question

A block on a frictionless table is connected as shown in FIGURE P15.75 to two springs having spring constants $k1$and $k2$. Find an expression for the block’s oscillation frequency $f$in terms of the frequencies $f1$and $f2$at which it would oscillate if attached to spring $1$or spring $2$alone.

Short answer

The expression for the block’s oscillation frequency$f=\sqrt{\frac{f_1^2{f}_2^2}{f_1^2+f_2^2}}$

Step-by-step solution

Introduction.

Newton's Second Law states that the net force $F$acting on a mass $m$body is proportional to the acceleration $a$of the body.

$\sum F=ma$

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

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

The Principles.

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:

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

 Principles.

In basic 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 $k$, not the amplitude.

The Given data and Required data.

Given data:

The following are the spring constants for the two springs:$k_1$and $k_2$

Figure $15.75$depicts a mass $m$block on a frictionless table attached to two springs.

If the block is only coupled to springs $1$or $2$, its oscillation frequencies are:$f_1$and $f_2$

Required data:

The oscillation frequency $f$of the block must be determined.

Solution.

Because the mass is exclusively in touch with spring $2$, it exerts a force on it $F_{\text{mass on}2}$. Furthermore, because spring $1$exerts a force according to Equation $\left(1\right)$, the net force on spring $2$must be zero.

$F_{\text{net on}2}=0$

$F_{\text{mass on}2}=F_{1\text{on}2}$

Using Hooke's law from Equation $\left(2\right)$, $F_{\text{mass on}2}$can be found:$F_{\text{mass on}2}=F_{1\text{on}2}=k_2\Delta x_2$

where $\Delta x_2$is the spring$2$ displacement.

Step 6

Spring $1$is similarly affected by the wall, which exerts a force of $F_{\text{wall on}1}$. In addition, spring $2$exerts a force $F_{2\text{on}1}$on $1$on spring $1$in the opposite direction as $F_{\text{wall on}1}$. Because spring $1$is likewise supposed to be massless, the net force exerted on it must be zero, according to Newton's second law. So

$F_{\text{net on}1}=0$

$F_{\text{wall on}1}=F_{2\text{on}1}$

Using Hooke's law from Equation $\left(2\right)$, we can find $F_{\text{wall on}1}$:

$F_{\text{wall on}1}=F_{2\text{on}1}=k_1\Delta x_1$

where $\Delta x_1$is the spring$1$displacement.

Step 7

The force $F_{2\text{on}1}$on $1$exerted by spring $2$on spring 1 is equal in size to the force $F_{1\text{on}2}$exerted by spring $1$on spring $2$according to Newton's third law. So

$F_{2\text{on}1}=F_{1\text{on}2}$

Substitute $F_{2\text{on}1}$from Equation $\left(5\right)$and $F_{1\text{on}2}$from Equation (4):

$k_2\Delta x_2=k_1\Delta x_1$

Step 8

The force $F_{\text{mass on}2}$exerted by the mass on spring $2$is also equivalent in size to the force $F_{2\text{on mass}}$exerted by spring $2$on the mass, according to Newton's third law.

$F_{\text{mass on}2}=F_{2\text{on mass}}=F_{\text{mass}}$

$F_{\text{mass}}=k_2\Delta x_2=k_1\Delta x_1$

Since spring $2$is the only spring exerting a force on the mass, we utilised role="math" localid="1650090962956" $F_{\text{mass}}$instead of $F_{2\text{on mass}}$.

Step 9

The length change of the complete system of two springs will be equal to the length change of each spring individually:

$\Delta x=\Delta x_1+\Delta x_2$

Substitute $\Delta x_1$and $\Delta x_2$in Equation $\left(6\right)$:

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

role="math" localid="1650091593875" $=\left(\frac{1}{k_1}+\frac{1}{k_2}\right)F_{\text{mass}}$

When we solve for $F_{\text{mass}}$,we get:

$F_{\text{mass}}=\left(\frac{1}{k_1}+\frac{1}{k_2}\right)\Delta x$

$=k_{\mathrm{eff}}\Delta x$

where $k_{\mathrm{eff}}=1/k_1+1/k_2$is the system's effective spring constant.

Step 10

Equation $\left(3\right)$gives the frequency of oscillation of the mass if it is only coupled to spring $1$:

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

Rearrange the equations and solve for $k_1$:

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

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

Step 11

Similarly, if the mass is just attached to spring $2$, the frequency of oscillation is:

Rearrange the equations and solve for $k_2$:

localid="1650092434459" $f_2^2=\frac{1}{4{\pi }^2}\frac{k_2}{m}$

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

Step 12

Equation $\left(3\right)$can also be used to calculate the frequency of oscillation of the mass and two springs in terms of the effective spring constant of the system:

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

$=\frac{1}{2\pi }\sqrt{\frac{k_1{k}_2}{\left(k_1+k_2\right)m}}$

Substitute $k_1$and $k_2$from Equations $\left(7\right)$and $\left(8\right)$:

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

$=\frac{1}{2\pi }\sqrt{\frac{16{\pi }^4{m}^2\left(f_1^2{f}_2^2\right)}{4{\pi }^2{m}^2\left(f_1^2+f_2^2\right)}}$

$=\frac{1}{2\pi }\sqrt{\frac{4{\pi }^2\left(f_1^2{f}_2^2\right)}{f_1^2+f_2^2}}$

$=\sqrt{\frac{f_1^2{f}_2^2}{f_1^2+f_2^2}}$