Question

A 1.30-m string of weight 0.0121N 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 equationy(x,t)=(8.50mm)cos(172rad?m?1x?2730rad?s?1t)Assume that the tension of the string is constant and equal to W.How much time does it take a pulse to travel the full length of the string?What is the weight W?How many wavelengths are on the string at any instant of time?What is the equation for waves traveling down the string?A. y(x,t)=(8.50mm)cos(172rad?m?1x?2730rad?s?1t)B. y(x,t)=(8.50mm)cos(172rad?m?1x+2730rad?s?1t)C. y(x,t)=(10.5mm)cos(172rad?m?1x+2730rad?s?1t)D. y(x,t)=(10.5mm)cos(172rad?m?1x?2730rad?s?1t)

Answer 1

Part a)

`t = 0.082 s`

Part b)

`W = 0.24 N`

Part c)

`N = 35.6`

Part d)

`y(x,t)=(8.50mm)cos(172rad/m x + 2730rad/s t)`

Step-by-step explanation:

As we know that

length of the string is L = 1.30 m

weight of the string is W = 0.0121 N

also we know that tension in the spring remains constant

and the Equation of wave in string is given as

`y(x,t)=(8.50mm)cos(172rad/m x - 2730rad/s t)`

Part a)

Speed of wave is given as

`v = (\omega)/(k)`

here we know that

`\omega = 2730 rad/s`

`k = 172 rad/s`

so we have

`v = (\omega)/(k)`

`v = (2730)/(172)`

`v = 15.9 m/s`so time taken by the wave to reach the top point is given as

`t = (L)/(v)`

`t = (1.30)/(15.9) = 0.082 s`

Part b)

Also we know that wave speed is given as

`v = \sqrt{(T)/(m/L)}`

here we have

`m = (0.0121)/(9.81)`

`m = 1.233* 10^(-3) kg`

also we have

`v = \sqrt{(W)/((1.233* 10^(-3))/1.30)`

`15.9 = \sqrt{(W)/((1.233* 10^(-3))/1.30)`

`W = 0.24 N`

Part c)

Wavelength of the wave travelling on the string

`\lambda = (2\pi)/(k)`

`\lambda = (2\pi)/(172)`

`\lambda = 0.0365 m`

so number of wavelengths on the string is given as

`N = (L)/(\lambda)`

`N = (1.30)/(0.0365)`

`N = 35.6`

Part d)

Since the direction of the wave is reversed when it travels back

so here we have equation of wave given as

`y(x,t)=(8.50mm)cos(172rad/m x + 2730rad/s t)`