A 1.60-m string of weight 0.0129 N is tied to the ceiling at its upper end, and

Question

A 1.60-m string of weight 0.0129 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(172radm1x2730rads1t)
Assume that the tension of the string is constant and equal to W.

Part A

How much time does it take a pulse to travel the full length of the string?

Part B

What is the weight W?

Part C

How many wavelengths are on the string at any instant of time?

Part D

What is the equation for waves traveling down the string?

What is the equation for waves traveling down the string?

y(x,t)=(8.50mm)cos(172radm1x2730rads1t) y(x,t)=(8.50mm)cos(172radm1x+2730rads1t) y(x,t)=(10.5mm)cos(172radm1x+2730rads1t) y(x,t)=(10.5mm)cos(172radm1x2730rads1t)

Explanation / Answer

given data
d=1.60m
weight = 0.0129N

(1) t = d/v where d = 1.6 m is the length of the string and v is the velocity of the wave traveling up the string

v = /k = (1/2730) / (1/172) = 0.063 m/s

t = (1.6) / (0.063) = 25.4 seconds

(2) v = sqrt(T/), where T is the tension in the string and is the linear mass density of the string

Here the tension in the string is created by the weight W, thus T = W, and = m/L, where m is the mass of the string and L is its length. Therefore,

v = sqrt(WL/m)

W = mv²/L = (0.0129/9.81)(0.063)^2 / (1.6) = 3.26 x 10^-6 N

(3) 1

(4) Same as the waves going up.

y(x,t)=(8.50mm)cos(172radm1x2730rads1t)

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