# A block is connected between two springs on a frictionless, horizontally level track. The other...

## Question:

A block is connected between two springs on a frictionless, horizontally level track. The other end of each spring is respectively anchored to a vertical pin set at opposite ends of the track. If the spring constants are determined to be 3 N/m and 1.6 N/m, and the period of oscillation for the block when slightly displaced from the equilibrium point is 0.6 s, what is the mass of the block?

## Spring Constant:

Whenever more than one spring connects with a specific body in a parallel fashion (connection), then the equivalent spring constant's value can be evaluated with an equation's help.

{eq}{k_{\rm{t}}} = {k_1} + {k_2} + {k_3} {/eq}.

Here, {eq}{k_{\rm{t}}} {/eq} represent the equivalent spring constant and {eq}{k_1} {/eq}, {eq}{k_2} {/eq} and {eq}{k_3} {/eq} are the individual spring constants of the springs.

The value of spring stiffness also depends on the spring's length it means if a spring of specific size cuts into parts, then the spring stiffness varies (increases).

## Answer and Explanation: 1

**Given Data**

- The spring constant of first spring is {eq}{k_1} = 3\;{\rm{N/m}} {/eq}.

- The spring constant of second spring is {eq}{k_2} = 1.6\;{\rm{N/m}} {/eq}.

- The time period of oscillation is {eq}T = 0.6\;\sec {/eq}.

As the one ends of both spring are attached with opposite vertical pin and other ends of both spring attached with the block so, this is the parallel combinations of both spring. Therefore the equivalent spring constant can be calculated as,

{eq}\begin{align*} {k_{{\rm{eq}}}} &= 3 + 1.6\\ &= 4.6\;{\rm{N/m}} \end{align*} {/eq}

The formula to calculate the mass of the block is given by,

{eq}T = {{2\pi }}\sqrt {\dfrac{{m}}{{k_{{\rm{eq}}}}}} {/eq}.

Here, {eq}m {/eq} is the mass of block.

Substitute all the known values in the above formula.

{eq}\begin{align*} T &= {{2\pi }}\sqrt {\dfrac{{m}}{{k_{{\rm{eq}}}}}} \\ 0.6 &= {{2\pi }}\sqrt {\dfrac{m}{{4.6}}} \\ m &\approx 0.042\;{\rm{kg}} \end{align*} {/eq}

Thus, the mass of the block is {eq}0.042\;{\rm{kg}} {/eq}.

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.