Question

The velocity of a particle performing S.H.M. are 0.13 m/s and 0.12 m/s. when it is at

0.12 m and 0.13 m from the mean position, respectively. Then, the amplitude is​

Answer 1

we know, velocity of particle = $\omega\sqrt{A^2-x^2}$

where, $\omega$ is angular frequency, A is amplitude and x is distance of particle from the mean position.

it is given that velocity of particle is 0.13m/s when it is at 0.12m from the mean position.

so, 0.13 = $\omega$√{A²-(0.12)²}.......(1)

again, velocity of particle is 0.12 m/s when it is at 0.13 m from the mean position.

so, 0.12=$\omega$√{A²-(0.13)²}......(2)

from equations (1) and (2),

0.13/0.12 = √{A² - (0.12)²}/√{A²-(0.13)²}

or, 13/12 = √{(10A)² - (12)²}/√{(10A)² - (13)²}

or, 13²/12² = {(10A)² - (12)²}/{(10A)² -(13)²}

or, 13²(10A)² - (13)⁴ = 12²(10A)² - (12)⁴

or, (13² - 12²)(10A)² = (13)⁴ - (12)⁴

or, (10A)² = (13² + 12²)

or, 100A² = (169 + 144)

or, 100A² = 313

or, A = √{313/100} = 1.76 m