Question

A wave traveling along a string is described by eq. y(x,t) =0.05 sin(40x-5t)
in which numerical constants are in SI UNITS. Calculate
A.Amplitude
B.Wavelength
C.Time period
D.Frequency

Answer 1

Answer:

Explanation:

hope it helps....

Answer 2

The amplitude is 0.05 m, the wavelength is 0.1571 m, the time period is 1.2566 s, and the frequency is 0.7958 Hz.

Step-by-step Explanation:

Given: The wave equation y(x,t) =0.05 sin(40x - 5t)

To Find: Amplitude, Wavelength, Time Period, Frequency

Solution:

• Amplitude

The general wave equation is y(x,t) = A sin(kx - ωt) . . . . . (1)

And, the given wave equation is y(x,t) = 0.05 sin(40x - 5t) . . . . . (2)

On comparing (1) and (2), we have Amplitude A = 0.05m

• Wavelength

On comparing (1) and (2), we have k = 40

Since, $k = \frac{2 \pi}{\lambda} \Rightarrow \lambda = \frac{2 \pi}{k}$

we get, $\lambda = \frac{2 \pi}{k} \Rightarrow \lambda = \frac{2 \times 3.14}{40} = 0.1571 m$

• Time Period

On comparing (1) and (2), we have ω = 40

Since, $\omega = \frac{2 \pi}{T} \Rightarrow T = \frac{2 \pi}{\omega}$

we get, $T = \frac{2 \pi}{\omega} \Rightarrow T = \frac{2 \times 3.14}{5} = 1.2566 s$

• Frequency

We know that $f = \frac{1}{T}$, therefore,

$f = \frac{1}{1.2566} = 0.7958 Hz$

Hence, the amplitude is 0.05 m, the wavelength is 0.1571 m, the time period is 1.2566 s, and the frequency is 0.7958 Hz.