A transverse wave propagating on a stretched string of linear density 3 - 10-4 kgm -1 is represented by the equation - y-0-2sin1-5x-60t where x is in metre and t is in secondsThe tension in the string in N

The correct option is B 0.48 Velocity v=√(Tμ) μ=3× 10^ 4 kgm^ 1 ∴ T=μ v^2 =3× 10^ 4× ( wk )^2 =3× 10^ 4× ( 601.5 )^2 =0.48N ...

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Standard XII

Physics

Sound Wave:Displacement Equation

A transverse ...

Question

A transverse wave propagating on a stretched string of linear density $3 \times 10^{- 4} k g m^{- 1}$ is represented by the equation :
$y = 0.2 sin (1.5 x + 60 t)$ where $x$ is in metre and $t$ is in seconds
The tension in the string in $N$

0.24

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0.48

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1.20

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1.80

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Solution

The correct option is B $0.48$
Velocity $v = \sqrt{\frac{T}{μ}}$
$μ = 3 \times 10^{- 4} k g m^{- 1}$
$∴ T = μ v^{2}$
$= 3 \times 10^{- 4} \times {(\frac{w}{k})}^{2}$
$= 3 \times 10^{- 4} \times {(\frac{60}{1.5})}^{2}$
$= 0.48 N$

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A transverse wave propagating on a stretched string of linear density 3×10−4kgm−1 is represented by the equation :y=0.2sin(1.5x+60t) where x is in metre and t is in secondsThe tension in the string in N

The correct option is B 0.48Velocity v=√Tμμ=3×10−4 kgm−1∴T=μv2=3×10−4×(wk)2=3×10−4×(601.5)2=0.48N

A transverse wave propagating on a stretched string of linear density $3 \times 10^{- 4} k g m^{- 1}$ is represented by the equation :
$y = 0.2 sin (1.5 x + 60 t)$ where $x$ is in metre and $t$ is in seconds
The tension in the string in $N$

The correct option is B $0.48$
Velocity $v = \sqrt{\frac{T}{μ}}$
$μ = 3 \times 10^{- 4} k g m^{- 1}$
$∴ T = μ v^{2}$
$= 3 \times 10^{- 4} \times {(\frac{w}{k})}^{2}$
$= 3 \times 10^{- 4} \times {(\frac{60}{1.5})}^{2}$
$= 0.48 N$