Question

The end point of a spring oscillates with a period of 2.0 swhen a block with mass mis attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Find m.

Short answer

Mass of the block is 1.6 kg.

Step-by-step solution

Step-by-Step Solution

• The period of the block,$T_1=2.0\mathrm{s}$.
• The new period after the increased mass,$T_2=3.0\mathrm{s}$ .
• The original mass of the block, m.
• The increased mass, m+2kg .

Understanding the concept of oscillations

The period T is the time required for one complete oscillation, or cycle. It is related to the frequency by

$\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathit{f}}$

It is also related to mass (m) and force constant (k) by the formula,
$\mathit{T}\mathbf{=}\mathbf{2}\mathit{\tau }\mathit{\tau }\sqrt{\frac{\mathit{m}}{\mathit{k}}}$

By using the formula for the period, we can write the expression for the period for different masses. Taking the ratio of these two equations we can find the mass of the block.

Formula:

The time period of oscillations,$T=2\tau \tau \sqrt{\frac{m}{k}}$ (i)

Calculation of mass, m

Using equation (i), the first period for the mass of the block is given as:

$T_1=2\tau \tau \sqrt{\frac{M_1}{k}}$

(ii)

After attaching the mass of 2 kg, the new time period using equation (i) can be written as:

$T_2=2\tau \tau \sqrt{\frac{M_2}{k}}$

(iii)

Taking the ratio of equation (ii) and (iii), we get

$$\begin{gathered} \frac{T_1}{T_2}=\frac{\sqrt{M_1}}{M_2} \\ \frac{2}{3}=\frac{\sqrt{m}}{\sqrt{m+2}} \\ \frac{4}{9}=\frac{m}{\left(m+2\right)}\left(\mathrm{squaring}\mathrm{both}\mathrm{the}\mathrm{sides}\right) \end{gathered}$$

$$\begin{gathered} 9m-4m=8 \\ 5m=8 \\ m=8/5 \\ =1.6\mathrm{kg} \end{gathered}$$

Hence, the value of mass of the block is 1.6 kg.