A string is producing transverse vibration whose equation is y=0.021 sin (x+30 t), Where x and y are in meters and t is in seconds. If the linear density of the string is 1.3 × 10-4 kg / m, then the tension in the string in N will be :

Question

A string is producing transverse vibration whose equation is y=0.021 sin (x+30 t), Where x and y are in meters and t is in seconds. If the linear density of the string is 1.3 × 10-4 kg / m, then the tension in the string in N will be : (A) 10 (B) 0.5 (C) 1 (D) 0.117

Tardigrade Admin · Accepted Answer

We use y=0.021 sin (x+30 t) Wave speed is given as v=(ω/k)=(30/1)=30 m / s Using, v=√(T/m) ⇒ 30=√(T/1.3 × 10-4) ⇒ T=0.117 N.

Q. A string is producing transverse vibration whose equation is $y = 0.021 sin (x + 30 t)$ , Where $x$ and $y$ are in meters and $t$ is in seconds. If the linear density of the string is $1.3 \times 1 0^{- 4} k g / m$ , then the tension in the string in $N$ will be :

10

0.5

1

0.117

We use
$y = 0.021 sin (x + 30 t)$
Wave speed is given as
$v = \frac{ω}{k} = \frac{30}{1} = 30 m / s$
Using, $v = \frac{T}{m}$
$\Rightarrow 30 = \frac{T}{1.3 \times 1 0 ^{- 4}}$
$\Rightarrow T = 0.117 N .$

Q. A string is producing transverse vibration whose equation is y=0.021sin(x+30t), Where x and y are in meters and t is in seconds. If the linear density of the string is 1.3×10−4kg/m, then the tension in the string in N will be :

10

0.5

1

0.117

We use y=0.021sin(x+30t) Wave speed is given as v=kω​=130​=30m/s Using, v=mT​​ ⇒30=1.3×10−4T​​ ⇒T=0.117N.

Q. A string is producing transverse vibration whose equation is $y = 0.021 sin (x + 30 t)$ , Where $x$ and $y$ are in meters and $t$ is in seconds. If the linear density of the string is $1.3 \times 1 0^{- 4} k g / m$ , then the tension in the string in $N$ will be :

We use
$y = 0.021 sin (x + 30 t)$
Wave speed is given as
$v = \frac{ω}{k} = \frac{30}{1} = 30 m / s$
Using, $v = \frac{T}{m}$
$\Rightarrow 30 = \frac{T}{1.3 \times 1 0 ^{- 4}}$
$\Rightarrow T = 0.117 N .$