The equation of a transverse wave travelling wave on a string is:

Y=3 cos[3.14(0.5x-200t)]
Where x and y in cm and t is in second.

a)find the amplitude, wave length,frequency period,velocity of propagation.

b)if the mass per unit length of the string is 0.5 kg/m find the tension.

To find the amplitude, wavelength, frequency, period, and velocity of propagation, we need to analyze the given equation:

Y = 3 cos[3.14(0.5x - 200t)]

a) Amplitude:
The amplitude of a wave is the maximum displacement of any particle in the medium from its equilibrium position. In this case, the amplitude is given as 3.

b) Wavelength:
The wavelength (λ) is the distance between two consecutive points on a wave that are in phase. Looking at the equation, we can see that the coefficient of x is 0.5. The wavelength is given by the formula:

λ = 2π/k, where k is the coefficient of x in the equation

In this case, k = 0.5, so:

λ = 2π/0.5 = 4π

c) Frequency:
The frequency (f) is the number of complete cycles of the wave passing a given point per unit time. It is measured in Hertz (Hz). The frequency is related to the angular frequency (ω) through the formula:

ω = 2πf, where ω is in radians per second

Comparing the equation to the standard form of a cosine function, we can see that the angular frequency is given by:

ω = 3.14 × 200 = 628

To find the frequency, we can divide the angular frequency by 2π:

f = ω/2π = 628/2π ≈ 100 Hz

d) Period:
The period (T) is the time it takes for one complete cycle of the wave to pass a given point. It is the reciprocal of the frequency:

T = 1/f = 1/100 ≈ 0.01 s

e) Velocity of Propagation:
The velocity of propagation (v) is the speed at which the wave is moving through the medium. It is given by the formula:

v = λf = wavelength × frequency

v = 4π × 100 = 400π cm/s (approximately)

b) Tension:
To find the tension in the string, we can use the wave equation:

v = √(T/μ), where T is the tension and μ is the mass per unit length of the string

The mass per unit length (μ) is given as 0.5 kg/m.

Rearranging the equation to solve for T:

T = μv^2

T = 0.5 × (400π)^2 ≈ 100,530 N

Therefore, the tension in the string is approximately 100,530 Newtons.