Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t - 9x) where distance and time are measured is SI units. The tension in the string is :

Question

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t - 9x) where distance and time are measured is SI units. The tension in the string is : (A) 10 N (B) 12.5 N (C) 7.5 N (D) 5 N

Tardigrade Admin · Accepted Answer

y =0.03 sin(450 t - 9x) v= (ω/k) =(450/9) =50 m/s v= √(T/μ) ⇒ (T/μ) =2500 ⇒ T = 2500 ×5 ×10-3 = 12.5 N

Q. Equation of travelling wave on a stretched string of linear density $5 g / m$ is $y = 0.03 s in (450 t - 9 x)$ where distance and time are measured is SI units. The tension in the string is :

10 N

12.5 N

7.5 N

5 N

$y = 0.03 sin (450 t - 9 x)$
$v = \frac{ω}{k} = \frac{450}{9} = 50 m / s$
$v = \frac{T}{μ} \Rightarrow \frac{T}{μ} = 2500$
$\Rightarrow T = 2500 \times 5 \times 1 0^{- 3}$
$= 12.5 N$

Q. Equation of travelling wave on a stretched string of linear density 5g/m is y=0.03sin(450t−9x) where distance and time are measured is SI units. The tension in the string is :

10 N

12.5 N

7.5 N

5 N

y=0.03sin(450t−9x) v=kω​=9450​=50m/s v=μT​​⇒μT​=2500 ⇒T=2500×5×10−3 =12.5N

Q. Equation of travelling wave on a stretched string of linear density $5 g / m$ is $y = 0.03 s in (450 t - 9 x)$ where distance and time are measured is SI units. The tension in the string is :

$y = 0.03 sin (450 t - 9 x)$
$v = \frac{ω}{k} = \frac{450}{9} = 50 m / s$
$v = \frac{T}{μ} \Rightarrow \frac{T}{μ} = 2500$
$\Rightarrow T = 2500 \times 5 \times 1 0^{- 3}$
$= 12.5 N$