A1.20-m string of weight 0.0121 n is tied to the ceiling at its upper end, and the lower end supports a weight w. neglect the very small variation in tension along the length of the string that is produced by the weight of the string. when you pluck the string slightly, the waves traveling up the string obey the equation y(x, t)=(8.50mm)cos(172rad⋅m−1x−2730rad⋅ s−1t)
assume that the tension of the string is constant and equal to w.

a. how much time does it take a pulse to travel the full length of the string?
b. what is the weight w?
c. how many wavelengths are on the string at any instant of time?

Explanation:

given wave = 8.5 x 10⁻³ cos ( 172 x - 2730t )

here wave no k = 172

angular frequency ω = 2730

velocity of wave on the string

a )

v = ω / k

= 2730 / 172

= 15.87 m/s

time taken to travel full length

= 1.2 / 15.87

= 75.6 x 10⁻³ s

b )

For velocity of wave on the wire the formula is

T is tension and m is mass per unit length of wire

m = .0121 / ( 9.8 x 1.2 )

= 1.03 x 10⁻³ kg / m

15.87 =

W = .259 N

c ) wavelength λ = 2π / k

= 2 x 3.14 / 172

= .0182 m

no of wave length

n = 1.2 / .0182

= 66 approx .

overcharge is the right answer system

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle