Question

ans is 1rad and 5m pleas solve​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

A body executing S.H.M.S.H.M. along a straight line has a velocity of ${\sf 3m/s}$ when it is at a distance of 4m from its mean position and ${\sf 4m/s}$ when it is at a distance of 3m from its mean position. Its angular frequency and amplitude are:

$\huge\underline{\underline{\bf \orange{Solution-}}}$

• ${\sf v_1=3\:m/s}$
• ${\sf x_1=4\:m}$
• ${\sf v_2=4\:m/s}$
• ${\sf x_2=3\:m/s}$

✪$\large{\boxed{\sf \blue{V=\omega\sqrt{A^2-X^2}}}}$

A = Amplitude

$\implies{\sf v_1=\omega\sqrt{A^2-x_1}}$

$\implies{\bf 3=\omega\sqrt{A^2-(4)^2}\:\:\:\:→(1)}$

$\implies{\sf v_2=\omega\sqrt{A^2-x_2} }$

$\implies{\bf 4=\omega\sqrt{A^2-(3)^2}\:\:\:→(2)}$

$\implies{\sf \dfrac{3}{4}=\dfrac{\omega}{\omega}\dfrac{\sqrt{A^2-(4)^2}}{A^2-(3)^2}} }$

$\implies{\sf \dfrac{9}{16}=\dfrac{A^2-16}{A^2-9}}$

$\implies{\sf 9(A^2-9)=16(A^2-16) }$

$\implies{\sf 9A^2-81=16A^2-256}$

$\implies{\sf 9A^2-16A^2=-256+81}$

$\implies{\sf -7A^2=-175}$

$\implies{\sf A^2=25}$

$\implies{\bf \red{Amplitude (A)=5\:m}}$

Putting Value of A in equation 1 we get ➝

$\implies{\sf 3=\omega\sqrt{A^2-(4)^2} }$

$\implies{\sf 3=\omega\sqrt{5^2-4^2} }$

$\implies{\sf 3=\omega\sqrt{25-16}}$

$\implies{\sf 3^2=\omega×9}$

$\implies{\bf \red{\omega = 1\:rad/s}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Amplitude ${\bf \red{5\:m}}$

Angular velocity ${\bf \red{\omega = 1\:rad/s}}$