Question

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 $4.9 \times 10^{-7} m^2$. 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 $n \times 10^9 N m^{-2}$, then the value of n is
A. 4
B. 2
C. 8
D. 1

Answer 1

Use the same technique for deriving the formula for frequency of simple harmonic equation as we use when there is a spring. The expression for force involved can be obtained from Hooke's law relation.
Formula used:
The frequency of oscillation can be written as
$\omega = \sqrt{ \dfrac{Y A}{m L} }$.

Complete answer:
The standard equation for a particle performing simple harmonic motion is of the form:
$\dfrac{d^2 x}{dt^2} + \omega^2 x = 0$.
We are looking for this type of equation to get the frequency.

By Hooke's law we have:
$Y = \dfrac{stress}{strain} = \dfrac{FL}{Ax}$
where x is the change in length, A is the area of cross section, L is total length of the wire and F is the force acting on the wire.
Using this, we can write
$F = - \dfrac{YA}{L} x$
where we have included a negative sign as direction of x is downwards.
We can now write:
$\dfrac{d^2 x}{dt^2} + \dfrac{YA}{mL} x= 0$
where we kept F = ma first and then divided both sides by m.
Comparing this expression with standard SHM expression, we may write:
$\omega = \sqrt{ \dfrac{Y A}{m L} }$
$\implies Y = \dfrac{\omega^2 m L}{A}$.
Keeping the given values in this, we get:
$Y = \dfrac{(140)^2 \times 0.1 \times 1}{4.9 \times 10^{-7}} = 4 \times 10^9 N m^{-2}$.

So, the correct answer is “Option A”.

Additional Information:
Simple Harmonic motion is a periodic oscillatory motion that a body is said to perform if its motion can be described by standard SHM equation (that we wrote previously). A horizontal spring (in absence of other forces) going back and forth performs SHM motion, a pendulum with small amplitude is also an example of SHM.

Note:
The derivation for standard SHM expression in case of spring starts from F = -kx. If one remembers this and the expression for force concerned with the wire, one will be easily able to derive the frequency. One can draw the analogy from the spring case and apply here.