The frequency of vibration of a sonometer wire is given by:
\(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\)
Calculating for initial tension:
\(f_1 = \frac{1}{2} \sqrt{\frac{6}{\mu}}\)
Calculating for new tension:
\(f_2 = \frac{1}{2} \sqrt{\frac{54}{\mu}}\)
Given:
\(f_2 - f_1 = 12\)
Ratio of frequencies:
\(\frac{f_1}{f_2} = \frac{1}{3}\)
Substituting values:
\(f_1 = 6 \, \text{Hz}\)
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: