The extension in a string, obeying Hooke's law is $x$ . The speed of wave in the stretched string is $v$ . If the extension in the string in increased to $1.5 x$ , the new the speed of wave is $\sqrt{\frac{x}{y}} v$ . Find $x + y$ .

3

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Solution

The correct option is B $5$
According to Hooke's Law,
$T \propto x$
Thus speed, $v \propto \sqrt{T} \propto \sqrt{x}$
Thus $\frac{v^{'}}{v} = \sqrt{\frac{1.5 x}{x}} \approx 1.22$
$⟹ v^{'} = 1.22 v$

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The extension in a string, obeying Hooke's law is x. The speed of wave in the stretched string is v. If the extension in the string in increased to 1.5x, the new the speed of wave is √xyv. Find x+y.

The correct option is B 5According to Hooke's Law, T∝xThus speed,v∝√T∝√xThus v′v=√1.5xx≈1.22⟹v′=1.22v

The extension in a string, obeying Hooke's law is $x$ . The speed of wave in the stretched string is $v$ . If the extension in the string in increased to $1.5 x$ , the new the speed of wave is $\sqrt{\frac{x}{y}} v$ . Find $x + y$ .

The correct option is B $5$
According to Hooke's Law,
$T \propto x$
Thus speed, $v \propto \sqrt{T} \propto \sqrt{x}$
Thus $\frac{v^{'}}{v} = \sqrt{\frac{1.5 x}{x}} \approx 1.22$
$⟹ v^{'} = 1.22 v$