Question

A 20 cm long string , having a mass 1.0g is fixed at both ends .the tension in the string is 0.5n .the string is set into vibrations using an, external vibrator of frequency 100 hz .find the separation

Answer 1

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Answer 2

Answer:

The separation is the 10 cm.

Explanation:

Given that,

Length l = 20 cm

Mass m = 1.0 g

Tension T = 0.5 N

Frequency f = 100 Hz

We know that,

The formula of velocity on stretched string is defined as:

$v =\sqrt{\dfrac{T}{\dfrac{m}{L}}}$

$v=\sqrt{\dfrac{0.5}{\dfrac{1.0\times10^{-3}}{20\times10^{-2}}}}$

$v = 10\ m/s$

The formula of the wave length is defined as:

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{10}{100}$

$\lambda= 10\ cm$

Hence, The separation is the 10 cm.