Question 446: Waves

Question

At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire

Accepted Answer

The fundamental frequency of a vibrating string is given by
$$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, $$
where
Ã¢ÂÂ¢Ã¢ÂÂ$ L $ is the vibrating length of the wire,
Ã¢ÂÂ¢Ã¢ÂÂ$ T $ is the tension in the wire, and
Ã¢ÂÂ¢Ã¢ÂÂ$ \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_{	ext{poles}} $) compared to the equator ($ g_{	ext{eq}} $), where $ g_{	ext{eq}} < g_{	ext{poles}} $. Therefore:
At the poles, the tension is $ T_{	ext{poles}} = m g_{	ext{poles}} $,
At the equator, the tension is $ T_{	ext{eq}} = m g_{	ext{eq}} $.
Since the frequency is proportional to $ \sqrt{T} $, using the same wire length at the equator would yield a lower frequency:
$$ f_{	ext{eq}} = \frac{1}{2L}\sqrt{\frac{m g_{	ext{eq}}}{\mu}} < f_{	ext{poles}} = \frac{1}{2L}\sqrt{\frac{m g_{	ext{poles}}}{\mu}}. $$
To have resonance with the same tuning fork (which has the frequency $ f_{	ext{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:
$$ \frac{1}{2L'}\sqrt{\frac{m g_{	ext{eq}}}{\mu}} = \frac{1}{2L}\sqrt{\frac{m g_{	ext{poles}}}{\mu}}. $$
Canceling common factors, we get:
$$ \frac{\sqrt{g_{	ext{eq}}}}{L'} = \frac{\sqrt{g_{	ext{poles}}}}{L}, $$
which leads to
$$ L' = L \sqrt{\frac{g_{	ext{eq}}}{g_{	ext{poles}}}}. $$
Since $ g_{	ext{eq}} < g_{	ext{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 Ã¢ÂÂ the vibrating length of the wire should be decreased.

Answer

should be same

Answer

should be increased

Answer

should be decreased

Answer

should be 3 times the original length

Waves

At the poles, a stretched wire of a given length vibrates in unison with a tuning fork. At the equator, for same setting to produce resonance with same fork, the vibrating length of wire

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