The fundamental frequency of a vibrating string is given by

$$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, $$

where

• $ L $ is the vibrating length of the wire,

• $ T $ is the tension in the wire, and

• $ \mu $ is the mass per unit length.

When the string is stretched by a hanging mass, the tension $ T $ is due to the weight of the mass, i.e.,

$$ T = m g, $$

where $ g $ is the acceleration due to gravity.

At the poles, the acceleration due to gravity is slightly higher ($ g_{\text{poles}} $) compared to the equator ($ g_{\text{eq}} $), where $ g_{\text{eq}} < g_{\text{poles}} $. Therefore:

At the poles, the tension is $ T_{\text{poles}} = m g_{\text{poles}} $,

At the equator, the tension is $ T_{\text{eq}} = m g_{\text{eq}} $.

Since the frequency is proportional to $ \sqrt{T} $, using the same wire length at the equator would yield a lower frequency:

$$ f_{\text{eq}} = \frac{1}{2L}\sqrt{\frac{m g_{\text{eq}}}{\mu}} < f_{\text{poles}} = \frac{1}{2L}\sqrt{\frac{m g_{\text{poles}}}{\mu}}. $$

To have resonance with the same tuning fork (which has the frequency $ f_{\text{poles}} $), we need to increase the fundamental frequency of the wire at the equator. One way to do this is by decreasing the vibrating length $ L $ because frequency is inversely proportional to length.

Let the new required vibrating length be $ L' $. For resonance, we equate the frequency at the equator with that at the poles:

$$ \frac{1}{2L'}\sqrt{\frac{m g_{\text{eq}}}{\mu}} = \frac{1}{2L}\sqrt{\frac{m g_{\text{poles}}}{\mu}}. $$

Canceling common factors, we get:

$$ \frac{\sqrt{g_{\text{eq}}}}{L'} = \frac{\sqrt{g_{\text{poles}}}}{L}, $$

which leads to

$$ L' = L \sqrt{\frac{g_{\text{eq}}}{g_{\text{poles}}}}. $$

Since $ g_{\text{eq}} < g_{\text{poles}} $, it follows that

$$ L' < L. $$

Thus, to resonate with the same tuning fork at the equator, the vibrating length of the wire must be decreased.

Answer: Option C – the vibrating length of the wire should be decreased.

Waves

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