Ncert Exercises & Solutions

When a mass \(m\) is connected individually to two springs \(S_1\) and \(S_2,\) the oscillation frequencies are \(\nu_1\) and \(\nu_2.\) If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be:

1.	\(v_2+v_2\)	2.	\(\sqrt{v_1^2+v_2^2}\)
3.	\(\left(\dfrac{1}{v_1}+\dfrac{1}{v_1}\right)^{-1}\)	4.	\(\sqrt{v_1^2-v_2^2}\)

Hint: \(\omega = 2\pi \nu = \sqrt{\frac{k}{m}}\)

Step 1: Find the frequency of the spring-mass system.

The angular frequency \(\omega\) of a mass-spring system is given by; \(\omega = 2\pi \nu = \sqrt{\frac{k}{m}}\)
For each individual spring, we have: \(\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}, \quad \nu_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}\)
where \(k_1\) and \(k_2\) are the spring constants of springs \(S_1\) and \(S_2\) respectively.

Step 2: Find the effective spring constant \(k_{\text{eff}}.\)
When two springs are connected in parallel, their effective spring constant is:
\(k_{\text{eff}} = k_1 + k_2\)

Step 3: Find the individual frequencies.

When the mass is connected to the springs individually:

\(\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}, \quad \nu_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}\)
Squaring both sides: \(\nu_1^2 = \frac{k_1}{4\pi^2 m}\Rightarrow k_1= 4\pi^2m\nu^2_1\)
\( \nu_2^2 = \frac{k_2}{4\pi^2 m}\Rightarrow k_2 = 4\pi^2m\nu^2_2\)
\(\Rightarrow k_\text{eff} = k_1 +k_2\)

\(\Rightarrow \nu_\text{eff}^2=\nu_1^2+\nu_2^2\)
Taking the square root we get;
\(\nu_{\text{eff}} = \sqrt{\nu_1^2 + \nu_2^2}\)
Hence, option (2) is the correct answer.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions