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Standard XII

Physics

Phase Difference

The linear de...

Question

The linear density of a vibrating string is $1.3 \times 10^{- 4} k g / m$ . The transverse wave propagating along the string is described by $y = 0.021 sin (x + 30 t)$ where $x$ is in meter and $t$ is in second.The tension in the string is

0.12 N

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0.48 N

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1.2 N

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4.8 N

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Solution

The correct option is C $0.12 N$
$y = 0.021 sin (x + 30 t)$
So, $k = 1$ and $ω = 30$
$v = \frac{ω}{k} = \frac{30}{1} m / s .$
$v = \sqrt{\frac{T}{μ}}$
$v^{2} = \frac{T}{μ}$
$\Rightarrow T = v^{2} μ$
$= 30 \times 30 \times 1.3 \times 10^{- 4}$
$= 1200 \times 10^{- 4}$
$= 0.12 N .$

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The linear density of a vibrating string is 1.3×10−4kg/m. The transverse wave propagating along the string is described by y=0.021sin(x+30t) where x is in meter and t is in second.The tension in the string is

The correct option is C 0.12Ny=0.021 sin(x+30 t)So, k=1 and ω=30v=ωk=301m/s.v=√Tμv2=Tμ⇒T=v2μ=30×30×1.3×10−4=1200×10−4=0.12N.

The linear density of a vibrating string is $1.3 \times 10^{- 4} k g / m$ . The transverse wave propagating along the string is described by $y = 0.021 sin (x + 30 t)$ where $x$ is in meter and $t$ is in second.The tension in the string is

The correct option is C $0.12 N$
$y = 0.021 sin (x + 30 t)$
So, $k = 1$ and $ω = 30$
$v = \frac{ω}{k} = \frac{30}{1} m / s .$
$v = \sqrt{\frac{T}{μ}}$
$v^{2} = \frac{T}{μ}$
$\Rightarrow T = v^{2} μ$
$= 30 \times 30 \times 1.3 \times 10^{- 4}$
$= 1200 \times 10^{- 4}$
$= 0.12 N .$