Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
[tex]A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}[/tex].....(I)
[tex]tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}[/tex]....(II)
Put the value of [tex]\omega=0.628\ rad/s[/tex] in equation (I) and (II)
[tex]A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}[/tex]
[tex]A=0.0198[/tex]
From equation (II)
[tex]tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}[/tex]
[tex]d=0.0023[/tex]
Put the value of [tex]\omega=3.14\ rad/s[/tex] in equation (I) and (II)
[tex]A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}[/tex]
[tex]A=0.0203[/tex]
From equation (II)
[tex]tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}[/tex]
[tex]d=0.0120[/tex]
Put the value of [tex]\omega=6.28\ rad/s[/tex] in equation (I) and (II)
[tex]A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}[/tex]
[tex]A=0.0209[/tex]
From equation (II)
[tex]tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}[/tex]
[tex]d=0.0257[/tex]
Put the value of [tex]\omega=9.42\ rad/s[/tex] in equation (I) and (II)
[tex]A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}[/tex]
[tex]A=0.0217[/tex]
From equation (II)
[tex]tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}[/tex]
[tex]d=0.0413[/tex]
Hence, This is the required solution.
