Question

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore 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) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)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 fulllength of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

Answer 1

Answer:

Explanation:

mass of string = .0125 / 9.8

= 1.275 x 10⁻³ kg

Length of string l = 1.5 m .

m = mass per unit length

= ( .1.275 / 1.5) x 10⁻³ kg/m

m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = $\sqrt{\frac{W}{m } }$

28.08 = $\sqrt{\frac{W}{.85\times10^{-3} } }$

788.48 =  W / .85 X 10⁻³

W = 670 x  10⁻³ N .

c ) wave length

wave length =2π  / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)