Question

6.
A transverse wave of amplitude 20 cm wavelength
50 cm and frequency 4 Hz is propagating over a taut
string in the negative x direction. The equation of this
wave will be best described by (where t is in s and
x is in m)
(1) y = 0.2 sin(4x - 871) m
(2) y = 0.2 sin(4x + 8ft) m
(3) y = 0.2 sin(21eX - 4rt) m
(4) y = 0.2 sin(2x + 47) m
The pre​

Answer 1

follow me

sorry I didn't know answer

mark me as brainliest

I hope it is useful

Answer 2

Concept:

A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance.

A periodic variable's amplitude is a gauge of its change over a single period.

The frequency of a repeated event is its number of instances per unit of time.

Given:

The amplitude of the wave is 20 cm.

The wavelength of the wave is 20 cm.

The frequency is 4 Hz.

Find:

The equation of the wave.

Solution:

Amplitude, A = 20cm= 0.2m

Wavelength, $\lambda=50cm=0.5m$

Frequency of the wave, $f=4Hz$

The general equation of the transverse wave is given as:

$y=Asin(kx+wt)$ where A is the amplitude.

Now,

$k=\frac{2\pi }{\lambda}$ and $w=2\pi f$

Therefore,

$y=Asin(kx+wt)\\y=Asin(\frac{2\pi }{\lambda}x+2\pi ft)$

$y=0.2sin(\frac{2\pi }{0.5}x+2\pi (4)t)\\y=0.2sin(4\pi x+8\pi t)$

The equation of the transverse wave is $y=0.2sin(4\pi x+8\pi t)$.

#SPJ5