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If the tension in the stretched string fixed at both ends is increased by 21%, the fundamental frequency is found to change by 15Hz, Then the

Option: 1

original frequency is 150Hz


Option: 2

velocity of the propagation of the transverse wave along the string increases by 5%


Option: 3

velocity of the propagation of the transverse wave along the string increases by 10%


Option: 4

fundamental wavelength on the string does not chnage


Answers (1)

best_answer

f \propto \sqrt{T},

\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}

\frac{f_1+15}{f_1}=\sqrt{\frac{1.21 T_1}{T_1}}=1.1

Solving, we get f1=150Hz

v \propto \sqrt{T}

\therefore \frac{V_2}{V_1}=\sqrt{\frac{T_2}{T_1}}

v_2=\left(\sqrt{\frac{1.21 T_1}{T_1}}\right) \cdot v_1=1.1v,

Hence, increases in v is 10%

\frac{\lambda}{2}=e \quad \Rightarrow \quad \therefore \quad \lambda=2 l

\therefore fundamnetal wavelength =2\lambda is unchanged

Posted by

manish painkra

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