A transverse periodic wave on a string with a linear mass density of $0.200$ kg/m is described by the following equation $y = 0.05 sin (420 t - 21.0 x)$ where $x$ and $y$ are in metres and $t$ is in seconds.

$32 N$

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$42 N$

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$66 N$

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$80 N$

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Solution

The correct option is D
$80 N$

$V = \sqrt{\frac{T}{μ}} = \frac{ω}{k} T = μ {(\frac{ω}{k})}^{2} ∴= (0.2) {(\frac{420}{21})}^{2} = 80 N$

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A transverse periodic wave on a string with a linear mass density of 0.200 kg/m is described by the following equation where x and y are in metres and t is in seconds.

Q. A transverse periodic wave on a string with a mass density of

0.200 k g / m

is described by the following equation

y = 0.05 sin (420 t - 21.0 x)

,
where x and y are in meters and t is in seconds.
The tension in the string is equal to:

The linear density of a vibrating string is $10^{- 4} k g / m$ . A transverse wave is propagating on the string, which is described by the equation y = 0.02 sin (x + 30 t), where x and y are in meters and time t in seconds. Then tension in the string is

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A transverse periodic wave on a string with a linear mass density of 0.200kg/m is described by the following equation y=0.05sin(420t−21.0x) where x and y are in metres and t is in seconds.

The correct option is D 80N V=√Tμ=ωkT=μ(ωk)2∴=(0.2)(42021)2=80N

A transverse periodic wave on a string with a linear mass density of $0.200$ kg/m is described by the following equation $y = 0.05 sin (420 t - 21.0 x)$ where $x$ and $y$ are in metres and $t$ is in seconds.

The correct option is D
$80 N$

$V = \sqrt{\frac{T}{μ}} = \frac{ω}{k} T = μ {(\frac{ω}{k})}^{2} ∴= (0.2) {(\frac{420}{21})}^{2} = 80 N$