Worked example 11.3: Block and two springs

Question: A block of mass $m=3 {\rm kg}$ is attached to two springs, as shown below, and slides over a horizontal frictionless surface. Given that the force constants of the two springs are $k_1= 1200 {\rm N/m}$ and $k_2= 400 {\rm N/m}$, find the period of oscillation of the system.

Answer: Let $x_1$ and $x_2$ represent the extensions of the first and second springs, respectively. The net displacement $x$ of the mass from its equilibrium position is then given by

$$x = x_1 + x_2.$$

Let $f_1 = k_1 x_1$ and $f_2=k_2 x_2$ be the magnitudes of the forces exerted by the first and second springs, respectively. Since the springs (presumably) possess negligible inertia, they must exert equal and opposite forces on one another. This implies that $f_1 = f_2$, or

$$k_1 x_1 = k_2 x_2.$$

Finally, if $f$ is the magnitude of the restoring force acting on the mass, then force balance implies that $f = f_1 = f_2$, or

$$f = k_{\rm eff} x = k_1 x_1.$$

Here, $k_{\rm eff}$ is the effective force constant of the two springs. The above equations can be combined to give

$$k_{\rm eff} = \frac{k_1 x_1}{x_1+x_2} = \frac{k_1}{1+ k_1/k_2} = \frac{k_1 k_2}{k_1+k_2}.$$

Thus, the problem reduces to that of a block of mass $m=3 {\rm kg}$ attached to a spring of effective force constant

$$k_{\rm eff} = \frac{k_1 k_2}{k_1+k_2} = \frac{1200\times 400}{1200+400} = 300 {\rm N/m}.$$

The angular frequency of oscillation is immediately given by the standard formula

$$\omega = \sqrt{\frac{k_{\rm eff}}{m}}=\sqrt{\frac{300}{3}} = 10 {\rm rad./s}.$$

Hence, the period of oscillation is

$$T = \frac{2 \pi}{\omega} = 0.6283 {\rm s}.$$