A string wave equation is given y = 0.002 sin (300t - 15x) and mass density is (μ=(0.1 kg/m)) . Then find the tension in the string

Question

A string wave equation is given y = 0.002 sin (300t - 15x) and mass density is (μ=(0.1 kg/m)) . Then find the tension in the string (A) 30N (B) 20N (C) 40N (D) 45N

Tardigrade Admin · Accepted Answer

In string wave speed is V=√(T/μ) T=mV2 T=.1×20×20=40 N V=(ω/k)=(300/15)=20 M/S

Q. A string wave equation is given y = 0.002 sin (300t - 15x) and mass density is $(μ = \frac{0.1 k g}{m})$ . Then find the tension in the string

30N

20N

40N

45N

In string wave speed is
$V = \frac{T}{μ}$
$T = m V^{2}$
$T = .1 \times 20 \times 20 = 40 N$
$V = \frac{ω}{k} = \frac{300}{15} = 20$ M/S

Q. A string wave equation is given y = 0.002 sin (300t - 15x) and mass density is (μ=m0.1kg​) . Then find the tension in the string

30N

20N

40N

45N

In string wave speed is V=μT​​ T=mV2 T=.1×20×20=40N V=kω​=15300​=20 M/S

Q. A string wave equation is given y = 0.002 sin (300t - 15x) and mass density is $(μ = \frac{0.1 k g}{m})$ . Then find the tension in the string

In string wave speed is
$V = \frac{T}{μ}$
$T = m V^{2}$
$T = .1 \times 20 \times 20 = 40 N$
$V = \frac{ω}{k} = \frac{300}{15} = 20$ M/S