The equation of a wave on a string of linear mass density 0.04 kgm- 1 is given by, y=0.02(m)sin [2 π ((t/0.04 ( s )) - (x/0.50 ( m )))] The tension in the string is (in 10- 2 N ):

Question

The equation of a wave on a string of linear mass density 0.04 kgm- 1 is given by, y=0.02(m)sin [2 π ((t/0.04 ( s )) - (x/0.50 ( m )))] The tension in the string is (in 10- 2 N ): (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From given equation -V=(λ /T)=(0.50/0.04) Now V=√(T/μ ) ⇒ T=v2μ =(12.5)2× 0.04 =6.25=625× 10- 2 N

Q. The equation of a wave on a string of linear mass density $0.04 k g m^{- 1}$ is given by,
$y = 0.02 (m) s in [2 π (\frac{t}{0.04 ( s )} - \frac{x}{0.50 ( m )})]$
The tension in the string is (in $1 0^{- 2} N$ ):

From given equation $- V = \frac{λ}{T} = \frac{0.50}{0.04}$
Now $V = \frac{T}{μ} \Rightarrow T = v^{2} μ$
$= (12.5)^{2} \times 0.04$
$= 6.25 = 625 \times 1 0^{- 2} N$

Q. The equation of a wave on a string of linear mass density 0.04kgm−1 is given by, y=0.02(m)sin[2π(0.04(s)t​−0.50(m)x​)] The tension in the string is (in 10−2N ):

From given equation −V=Tλ​=0.040.50​ Now V=μT​​⇒T=v2μ =(12.5)2×0.04 =6.25=625×10−2N

Q. The equation of a wave on a string of linear mass density $0.04 k g m^{- 1}$ is given by,
$y = 0.02 (m) s in [2 π (\frac{t}{0.04 ( s )} - \frac{x}{0.50 ( m )})]$
The tension in the string is (in $1 0^{- 2} N$ ):

From given equation $- V = \frac{λ}{T} = \frac{0.50}{0.04}$
Now $V = \frac{T}{μ} \Rightarrow T = v^{2} μ$
$= (12.5)^{2} \times 0.04$
$= 6.25 = 625 \times 1 0^{- 2} N$