JEE PYQ: Oscillations Question 25
Question 25 - 2020 (06 Sep 2020 Shift 1)
An object of mass $m$ is suspended at the end of a massless wire of length $L$ and area of cross-section $A$. Young modulus of the material of the wire is $Y$. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:
(1) $f = \frac{1}{2\pi}\sqrt{\frac{mL}{YA}}$
(2) $f = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}$
(3) $f = \frac{1}{2\pi}\sqrt{\frac{mA}{YL}}$
(4) $f = \frac{1}{2\pi}\sqrt{\frac{YL}{mA}}$
Show Answer
Answer: (1)
Solution
Wire acts as spring with $k = \frac{YA}{L}$. $f = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}$.
BETA
