Question 246: Waves

Question

At 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

Step 1. Recall wire vibration frequency.
For a stretched string/wire fixed at both ends:
$ f = \frac{n}{2L} \sqrt{\frac{T}{\mu}} $
where
$f$: frequency,
$n$: harmonic number,
$L$: length between supports,
$T$: tension,
$\mu$: mass per unit length.
Step 2. What changes between poles and equator?
Tension in the wire is due to hanging mass providing weight.
$ T = Mg_{	ext{eff}} $
But effective gravity differs:
At poles: $g_{	ext{eff}} = g$.
At equator: reduced because of earthÃ¢ÂÂs rotation: $g_{	ext{eff}} = g - \omega^2R$.
Thus, effective gravity is
less at equator
.
So for same suspended mass, tension $T$ is smaller at equator.
Step 3. Effect on frequency.
If same length, tension is smaller, so frequency is smaller:
$ f \propto \sqrt{T}. $
At equator $f$ of the wire < fork frequency.
Step 4. To restore resonance.
Since $f \propto 	frac{1}{L}$, to
increase
the wireÃ¢ÂÂs frequency to match the same tuning fork, one must
decrease the vibrating length L
.
Ã¢ÂÂ Correct answer:
Option A: should be decreased.

Answer

should be decreased.

Answer

should be increased.

Answer

should be same.

Answer

should be three times.

Waves

At 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

Choose the correct answer: