Question 373: Oscillations

Question

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its crosssectional area is 4.9 $$ 	imes $$ 10
-7
m
2
. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s
−1
. If the Young’s modulus of the material of the wire is n $$ 	imes $$ 10
9
Nm
-2
, the value of n is

Accepted Answer

When mass m is pulled by a force F, the wire elongation x, length l, cross-sectional area A, and Young's modulus of wire material Y are related by $$Y = {{F/A} \over {x/l}}$$ i.e.,
$$F = (Y\,A/l)x$$.
The restoring force by the wire is equal but opposite to F i.e., F
r
= $$-$$F. Apply Newton's second law to get
$$m{d^2}x/d{t^2} =  - (YA/l)x =  - {\omega ^2}x$$.
This equation represents SHM with an angular frequency $$\omega  = \sqrt {YA/(lm)} $$. Substitute the values to get $$Y = {\omega ^2}lm/A = 4 	imes {10^9}$$ N/m
2
.

Oscillations

Integer / Numeric Answer Type