Question

One end of a long metallic wire of length $l$ is tied to the ceiling. The other end is tied to a massless spring of spring constant $K$. A mass ($M$) hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are $a$ and $Y$ respectively. If the mass is slightly pulled down and released, it will oscillate with a frequency $f$ equal to

• A. $\frac{1}{2\pi }\sqrt{\frac{Kl}{MYa}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{Ya}{Ml}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{YaK}{M(Ya+Kl)}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{K}{M}}$

Answer 1

The correct option is C $\frac{1}{2\pi }\sqrt{\frac{YaK}{M(Ya+Kl)}}$


We can consider it as two springs connected in series.
$K_{eff}=\frac{K_1{K}_2}{K_1+K_2}$
$K_1=K;$
& For string $K_2=\frac{Ya}{l}$

$\therefore$ The effective spring constant
$K_{eff}=\frac{K_1{K}_2}{K_1{K}_2}=\frac{K\left(\frac{Ya}{l}\right)}{K+\left(\frac{Ya}{l}\right)}=\frac{KYa}{Kl+Ya}$

Therefore, frequency of oscillation is
$n=\frac{1}{2\pi }\sqrt{\frac{K_{eff}}{M}}$
$n=\frac{1}{2\pi }\sqrt{\frac{KYa}{M(Kl+Ya)}}$