string tied to the ceiling
A 1.30m- string of weight 0.0122N 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) = (8.50 mm cos 172 rad . m^-1 (x) - 2730rad . s^-1 (t) )
Assume that the tension of the string is constant and equal to W.
1. How much time does it take a pulse to travel the full length of the string?
2. What is the weight W ?
unit = N
3. How many wave lengths are on the string at any instant of time?
4. What is the equation for waves traveling down the string?
( choose one)
A. y (x,t) = (8.50 mm cos 172 rad . m^-1 (x) - 2730rad . s^-1 (t) )
B. y (x,t) = (8.50 mm cos 172 rad . m^-1 (x) + 2730rad . s^-1 (t) )
C. y (x,t) = (10.50 mm cos 172 rad . m^-1 (x) + 2730rad . s^-1 (t) )
D. y (x,t) = (10.50 mm cos 172 rad . m^-1 (x) - 2730rad . s^-1 (t) )
Question 2 should be 0.241N
- 天同Lv 78 years agoFavorite Answer
1. 172 = 2.pi/入 where 入 is the wavelength
hence, 入 = 0.0365 m
and 2.pi.f = 2730 where f is the frequency of the wave
i.e. f = 434.5 Hz
Speed of wave = 0.0365 x 434.5 m/s = 15.86 m/s
Time taken for the wave to travel through the string
= 1.3/15.86 s = 0.082 s
2. Mass of string = 0.0122/g = 0.00122 kg
where g is the acceleration due to gravity, taken to be 10 m/s^2
Mass per unit length = 0.00122/1.3 kg/m = 9.385 x 10^-4 kg/m
Hence, 0.082^2 = W/9.385x10^-4
W = 6.31 x 10^-6 N
3. No. of wavelengths = 1.3/入 = 1.3/0.0365 = 35.6
4. option B