Two sinusoidal waves of the same frequency are to be sent in the same direction along a taut string. One wave has amplitude of,5.0 mm the other.8.0 mm (a) What phase difference${\mathbf{\varphi }}_{\mathbf{1}}$between the two waves results in the smallest amplitude of the resultant wave? (b) What is that smallest amplitude? (c) What phase difference${\mathbf{\varphi }}_{\mathbf{2}}$results in the largest amplitude of the resultant wave? (d) What is that largest amplitude? (e) What is the resultant amplitude if the phase angle is ($\left({\varphi }_1-{\varphi }_2\right)\mathbf{/}\mathbf{2}$)?

ii) Amplitude of wave two,y = 5.0mm

Consider two sinusoidal waves traveling with the same frequency in the same direction along a string. The amplitude of the resultant wave can be calculated by using the values of wave amplitude and the phase difference between the waves. The least amplitude between two sinusoidal wave’s results when the phase difference isand the largest amplitude is produced when the phase difference is 0. Also, the resultant amplitude can be calculated by using the Pythagoras theorem.

Formula:

Pythagoras theorem of hypotenuse of a right-angled triangle, $\mathrm{z}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$ (i)

The smallest amplitude resultant wave always occurs when the phase difference is ${\mathrm{\phi }}_1=\mathrm{\pi rad}$.This is because when the phase difference is $\pi$, the amplitude vectors would be directed opposite to each other which results in the least resultant amplitude.

Hence, the value of phase is$\pi \mathrm{rad}$

To calculatethesmallest amplitude, we can take the difference between the two amplitudes as they would be directed opposite to each other with a phase difference equal to$\mathrm{\pi }$as described above.

Therefore,the smallest amplitude is given as:

(0.8) -(5.0) = 3.0 mm

Hence,

the value f the amplitude is 3.0mm

To calculate phase difference${\mathrm{\phi }}_2$ the largest amplitude resultant wave always occurs whenthe phase difference${\mathrm{\phi }}_2=0\mathrm{rad}$

Hence, the value of phase is 0.

To calculatethelargest amplitude, we can add the two amplitudes as they would be directed in the same direction astheangle between them would be zero. Therefore, the largest amplitude is given as:

$\left(8.0\right)+\left(0.5\right)=13\mathrm{mm}$

Hence,

the value of largest amplitude is 13mm

Whenthe phase difference between them is

$$\begin{gathered} \left({\mathrm{\phi }}_1-{\mathrm{\phi }}_{}\right)/2=\left(\pi -0\right)/2 \\ =\pi /2 \end{gathered}$$

it implies that the two waves are perpendicular to each other. Hence, using equation (i) to get the resultant amplitude as follows:

$\sqrt{{\left(8.0\right)}^2+{\left(5.0\right)}^2}=9.4\mathrm{m}\mathrm{m}$

Hence,

the value of resultant amplitude is 9.4 mm