Question

IF THE THICkNESS OF A VIBRATING STRINGS IS INCREASED ITS PITCH​

Answer 1

For a vibrating string,

T (time period) = $2l\sqrt{\frac{\pi r^{2}d}{ Tension} }$

So, f (frequency) = 1/T => $\frac{1}{2l} \sqrt{\frac{ Tension}{\pi r^{2}d} }$

We know that pitch is directly proportional to Frequency.

Since $\pi r^{2} = area$

So, f(frequency) = $\frac{1}{2l} \sqrt{\frac{Tension}{area * density} }$

So, f is inversely proportional to $\sqrt{area}$

So, if the thickness increases, then the area of cross section increases hence, the frequency reduces.

If the frequency reduces, the pitch reduces.

Hence, if the thickness of vibrating string is increased (area of cross section is increases), the pitch reduces or the sound becomes more low pitched.