Question

Fill in the Blank : A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is $4.9\times10^{-7} m^{2}$. If the mass is pulled a little in the vertically downward direction and released, it performs SHM of angular frequency 140 rad/s. If the Young’s modulus of the material of the wire is $n\times10^{9} N/m^{2}$, the value of n is . . . . . .

Answer 1

we have not given maximum extension of wire . as particle attached to wire performs SHM so, it is same as amplitude of shm.

but here a good news , we have given given angular frequency.

force applied on object = weight of object

or, $m\omega^2A=mg$

where $\omega$ is angular frequency and i.e., 140 rad/s

so, now, $\omega^2A=g$

or, 140² × A = 10

or, A = 10/19600 = 1/1960 m

Young's modulus = stress/strain

= force applied × l/Area × A

= mg × l/area × A

given, m = 0.1kg, area = 4.9 × 10^-7 m², l = 1m and A = 1/1960 m

so, Young's modulus = 0.1 × 10 × 1/(4.9 × 10^-7 × 1/1960)

= 1960 × 10^7/4.9

= 4 × 10^9 N/m²

comparing with n × 10^9 N/m²

n = 4