The frequency of vibration of a given string is directly proportional to the square root of the tension T & inversely proportional to the length & to the diameter d. If T is increases by 50% & its halved, calc. the % change in the frequency.

Let $f$ be the frequency, $T$ be the tension, $L$ be the length, and $d$ be the diameter.
The problem states that the frequency $f$ is directly proportional to the square root of the tension $T$, and inversely proportional to the length $L$ and the diameter $d$. This can be written as:

$$f = \frac{k \sqrt{T}}{Ld}$$

where $k$ is the constant of proportionality.

Let the initial values be $f_1, T_1, L_1, d_1$.
The initial frequency is:

$$f_1 = \frac{k \sqrt{T_1}}{L_1 d_1}$$

The new values are:

• Tension $T$ is increased by 30%: $T_2 = T_1 + 0.30 T_1 = 1.30 T_1$
• Length $L$ is halved: $L_2 = \frac{1}{2} L_1 = 0.5 L_1$
• Diameter $d$ is halved: $d_2 = \frac{1}{2} d_1 = 0.5 d_1$

Substitute the new values into the relationship:

$$f_2 = \frac{k \sqrt{T_2}}{L_2 d_2}$$ $$f_2 = \frac{k \sqrt{1.30 T_1}}{(0.5 L_1)(0.5 d_1)}$$ $$f_2 = \frac{k \sqrt{1.30} \sqrt{T_1}}{0.25 L_1 d_1}$$

Rearrange the terms to relate $f_2$ to $f_1$:

$$f_2 = \left(\frac{\sqrt{1.30}}{0.25}\right) \left(\frac{k \sqrt{T_1}}{L_1 d_1}\right)$$

Since $f_1 = \frac{k \sqrt{T_1}}{L_1 d_1}$, we have:

$$f_2 = \frac{\sqrt{1.30}}{0.25} f_1$$

Calculate the numerical factor:

$$\frac{\sqrt{1.30}}{0.25} \approx \frac{1.140175}{0.25} \approx 4.5607$$

So, $f_2 \approx 4.5607 f_1$.

The percentage change is given by:

$$\text{Percentage Change} = \frac{f_2 - f_1}{f_1} \times 100\%$$

Substitute $f_2 \approx 4.5607 f_1$:

$$\text{Percentage Change} = \frac{4.5607 f_1 - f_1}{f_1} \times 100\%$$ $$\text{Percentage Change} = \frac{(4.5607 - 1) f_1}{f_1} \times 100\%$$ $$\text{Percentage Change} = 3.5607 \times 100\%$$ $$\text{Percentage Change} = 356.07\%$$

Since the value is positive, it is a percentage increase.

The percentage change in the frequency is approximately 356.1% increase.

$$356.1\% \text{ increase}$$