Question

The vibration of a string is represented by y=5sin(πx/15)cos(48nt), where x and y are in cm
and t in second then amplitude of the superposing
wave is
(1) 125 cm
(2) 5 cm
(3) 10 cm
(4) 2.5 cm​

Answer 1

The amplitude of the superposing  wave is 2.5 cm.

(4) is correct option.

Explanation:

Given that,

The vibration of a string is represented by

$y=5\sin(\dfrac{\pi x}{15})\cos(48 \pi t)$

Here, x and y in cm and t in second

Amplitude :

Amplitude is equal to the maximum displacement of the wave.

We need to calculate the amplitude of the superposing  wave

Using given equation

$y=5\sin(\dfrac{\pi x}{15})\cos(48 \pi t)$

Multiply and divided by 2

$y=\dfrac{5}{2}(\sin(\dfrac{\pi x}{15})\cos(40\pi t))\times2$

$y=\dfrac{5}{2}[\sin(\dfrac{\pi x}{15}+40\pi t)+\sin(\dfrac{\pi x}{15}-40\pi t)]$

Here, $y_{1}=\dfrac{5}{2}[\sin(\dfrac{\pi x}{15}+40\pi t)]$

$y_{2}=\dfrac{5}{2}[\sin(\dfrac{\pi x}{15}-40\pi t)]$

Now, the general equation is

$y=a\sin(\omega t-\dfrac{2\pi}{\lambda}+\alpha)$....(II)

Comparing equation (I) and (II)

So, The amplitude is

$A = \dfrac{5}{2}=2.5\ cm$

Hence, The amplitude of the superposing  wave is 2.5 cm.

Learn more :

Topic : amplitude of the wave

https://brainly.in/question/13677845

Answer 2

Explanation:

Hope this helps you

Thank you ❤️❤️