91Ó°ÊÓ

Wavelength of each string \(\lambda = 0.5\;{\rm{m}}\).

The mass per unit length \(\mu = 0.1\;{\rm{kg/m}}\).

Tension is \(F = 3.125\;N\)

And gravity due to acceleration is \(g = {\pi ^2}\).

The maximum acceleration is written as\({a_{\max }} = {\omega ^2}A = g\). The angular velocity is\(\omega = \frac{{2\pi v}}{\lambda }\). And the velocity is\(v = \sqrt {\frac{F}{\mu }} \).Thus, the minimum amplitude is given by:

\({A_{\min }} = \frac{{g{\lambda ^2}\mu }}{{4F}}\)

Here, \(F\) is tension, \(\mu \) is linear density .

According to the question,

\(\begin{array}{c}{A_{\min }} = \frac{{g{\lambda ^2}\mu }}{{4{\pi ^2}F}}\\ = \frac{{{\lambda ^2}\mu }}{{4F}}\\ = \frac{{0.5\;{\rm{m}} \times 0.5\;{\rm{m}} \times 0.1\;{\rm{kg/m}}}}{{3.125\;{\rm{N}}}}\\ = 2 \times {10^{ - 3}}\;{\rm{m}}\\ = 2\;{\rm{mm}}\end{array}\)

Hence, the minimum amplitude is \(2\;{\rm{mm}}\).