Answer in English on related scenario

The speed $V$ of a transverse wave on a stretched string is given by:

The speed of the transverse wave is $\boxed{200 \text{ m/s}}$.

For a string fixed at both ends, the fundamental wavelength ($\lambda_1$) is twice the length of the string:

The fundamental wavelength is $\boxed{1.32 \text{ m}}$.

The fundamental frequency ($f_1$) is related to the wave speed $V$ and fundamental wavelength $\lambda_1$ by the formula $f_1 = \frac{V}{\lambda_1}$:

The fundamental frequency is $\boxed{151.52 \text{ Hz}}$.

The fundamental frequency ($f_1$) of a wave on a string is given by the formula:

From this, we see that $f_1$ is directly proportional to the square root of the tension: $f_1 \propto \sqrt{T}$.

If the fundamental frequency is increased by 10%, the new frequency $f_1'$ will be $1.10 f_1$.
Thus, we can write:

Using the proportionality $f_1 \propto \sqrt{T}$:

Where $T'$ is the new tension.

Square both sides:

This means $T' = 1.21 T$.
The percentage increase in tension is:

The tension must be increased by $\boxed{21\%}$.

The formula for sound intensity level ($B$) is:

Where $I$ is the sound intensity, and $I_0$ is the reference intensity, which is $10^{-12} \text{ W/m}^2$.

We are given $B = 100 \text{ dB}$.

Divide both sides by 10:

To remove the logarithm, raise both sides to the power of 10:

Solve for $I$:

The sound intensity is $\boxed{10^{-2} \text{ W/m}^2}$.

The speed of sound in air is $340 \text{ m/s}$. We assume the emitted sound wave has the same frequency as the fundamental frequency of the string, which was calculated in part b) as $f_1 \approx 151.52 \text{ Hz}$.
Given:
Speed of sound in air, $V_{sound} = 340 \text{ m/s}$
Frequency of emitted sound, $f = 151.52 \text{ Hz}$

The relationship between wave speed ($V$), frequency ($f$), and wavelength ($\lambda$) is $V = f\lambda$.
Rearranging for wavelength:

The wavelength of the emitted sound wave is $\boxed{2.243 \text{ m}}$.