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Speed of a transverse wave on a stretched string. The wave velocity through a medium depends on its inertial and elastic properties. So the transverse wave through a stretched string is determined by two factors:
- Tension T in the string is a measure in the string. Without tension no can propagate in the string. of elasticity disturbance Dimensions of T = [Force] = [MLT 2]
- Mass per unit length or linear mass density m of the string so that the string can store kinetic energy.
Dimensions of {tex}m=\frac{[\text { Mass }]}{[\text { Length }]}=\left[\mathrm{ML}^{-1}\right]{/tex}
Now, dimensions of ratio {tex}\frac{T}{m}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{ML}^{-1}\right]}=\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]{/tex}
As the speed v has the dimensions [LT-1] so we can express v in terms of T and m as {tex}v=C \sqrt{\frac{T}{m}}{/tex}
From detailed mathematical analysis! or from experiments, the dimensionless constant C = 1. Hence the speed of transverse waves on a stretched string is given by
{tex}v=\sqrt{\frac{T}{m}}{/tex}
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Preeti Dabral 2 years, 7 months ago
Let the required ratio be k:1.
Then, by the section formula, the coordinates of P are
{tex}P \left( \frac { 4 k - 3 } { k + 1 } , \frac { - 9 k + 5 } { k + 1 } \right){/tex}
{tex}\therefore \quad \frac { 4 k - 3 } { k + 1 } = 2 \text { and } \frac { - 9 k + 5 } { k + 1 } = - 5{/tex} [{tex}\because{/tex} P(2, 5) is given]
{tex}\Rightarrow{/tex} 4k - 3 = 2k + 2 and -9k + 5 = -5k - 5
{tex}\Rightarrow{/tex} 2k = 5 and 4k = 10
{tex}\Rightarrow{/tex} {tex}k = \frac { 5 } { 2 }{/tex} in each case.
So, the required ratio is {tex}\frac { 5 } { 2 } : 1, {/tex} which is 5:2
Hence, P divides AB in the ratio 5:2.
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Preeti Dabral 2 years, 7 months ago
The range of movements of joints is called flexibility.
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