Question

A uniform spring with k = 8600 N.mis cut into pieces 1and 2of unstretched lengths${\mathit{L}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ and${\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{cm}$. What are (a)${\mathit{k}}_{\mathbf{1}}$and (b)${\mathit{k}}_{\mathbf{2}}$? A block attached to the original spring as in Fig.15-7oscillates at 200 Hz. What is the oscillation frequency of the block attached to (c) piece 1and (d) piece 2?

Short answer

a) The spring constant $k_1$is 22,886 N/m .

b) The spring constant $k_2$is 14,620 N/m .

c) Oscillation frequency to piece 1 is 312 Hz .

d) Oscillation frequency to piece 2 is 260 Hz .

Step-by-step solution

The given data

• Original spring constant,k=8600 N/m .
• The length of two pieces, ${\mathrm{L}}_1=7.0\mathrm{cm}\mathrm{and}{\mathrm{L}}_2=10.0\mathrm{cm}$.
• Original frequency of oscillations,f =200 Hz .

Understanding the concept of oscillation frequency

When spring is cut into pieces, the spring constants of the pieces are inversely proportional to the length of the piece. Using the formula for the resultant force constant for the series combination of the spring, we can calculate the force constants and new frequencies of the springs.

Formula:

The spring constant relation to the length of the spring,$k\alpha \frac{1}{L}$ (i)

The angular frequency of the spring,

$$\begin{gathered} \omega =2\tau \tau f \\ =\sqrt{\frac{k}{m}} \end{gathered}$$ (ii)

Here,

$\omega$is angular frequency

$f$is frequency

$k$is force constant

$m$is mass

a) Calculation of k1

Using condition (i), we get the spring constants of two pieces as:

So, we can write,$k_1\alpha \frac{1}{L_1}$ and$K_2\alpha \frac{1}{L_2}$

Taking ratio, we get

$$\begin{gathered} \frac{k_2}{k_1}=\frac{L_1}{L_2} \\ =\frac{7.0}{10.0} \\ {k}_2=0.7k_1.....................\left(1\right) \end{gathered}$$

Now, when the two pieces were connected together end to end, the arrangement is called as a series. Hence, the three spring constants (original and two new) are related as

$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$

Substituting the value of equation (1), we get the value of spring constant of piece 1 as:
role="math" localid="1657274770042" $$\begin{gathered} \frac{1}{k}=\frac{1}{k_1}+\frac{1}{0.7k_1} \\ =\frac{1.7k}{0.7k_1} \end{gathered}$$

Thus, we get,

$$\begin{gathered} k_1=\frac{1.7k}{0.7} \\ =\frac{1.7\times 8600}{0.7} \\ =20,886\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the value of the spring constant is 20,886 N/m .

b) Calculation of k2

And the spring constant from equation (1), we get the spring constant of piece 2 as:

$$\begin{gathered} k_2=0.7k_1 \\ =14,620\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the value of the spring constant is 14,620 N/m .

c) Calculation of frequency of block when attached to piece 1

From equation (ii), the frequency of oscillations of the block when attached to piece 1 and 2 can be given as:

$$\begin{gathered} f_1=\frac{1}{2\tau \tau }\sqrt{\frac{k_1}{m}} \\ {f}_2=\frac{1}{2\tau \tau }\sqrt{\frac{k_2}{m}} \end{gathered}$$

Taking ratio with respect to the original frequency of oscillations, we get

$$\begin{gathered} \frac{f_1}{f}=\sqrt{\frac{k_1}{m}\times \frac{m}{k}} \\ =\sqrt{\frac{k_1}{k}}..............................................\left(a\right) \end{gathered}$$

Hence, we determine the frequency of piece 1 as:
$$\begin{gathered} f_1=f\sqrt{\frac{k_1}{k}} \\ =200\sqrt{\frac{1.7}{0.7}} \\ =312\mathrm{Hz} \end{gathered}$$

Hence, the required value of frequency is 312 Hz .

d) Calculation of frequency of the block when attached to piece 2

Similarly, the frequency ratio for piece 2 using equation (a) can be given as:

$$\begin{gathered} \frac{f_2}{f}=\sqrt{\frac{k_1}{k}} \\ {f}_2=f\sqrt{\frac{k_2}{k}} \\ =200\sqrt{1.7} \\ =260\mathrm{Hz} \end{gathered}$$

Hence, the value of required frequency is 260 Hz .