Question

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m with damping characterized by y = 0.628 (1/s). If Fo = 0.5 N, calculate A and d for the following values of w: 0.628 rad/s, 3.14 rad/s, 6.28 rad/s, and 9.42 rad/s.

Answer 1

Step-by-step 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

`A=\frac{(F_(0))/(m)}{\sqrt{(\omega_(0)^2-\omega^(2))^2+y^2\omega^2}}`.....(I)

`tan d=(y\omega)/((\omega^2-\omega^2))`....(II)

Put the value of
`\omega=0.628\ rad/s` in equation (I) and (II)

`A=((0.5)/(0.254))/(√((10.0^2-0.628)^2+0.628^2*0.628^2))`

`A=0.0198`

From equation (II)

`tan d=(0.628*0.628)/(((10.0^2-0.628)^2))`

`d=0.0023`

Put the value of
`\omega=3.14\ rad/s` in equation (I) and (II)

`A=((0.5)/(0.254))/(√((10.0^2-3.14)^2+0.628^2*3.14^2))`

`A=0.0203`

From equation (II)

`tan d=(0.628*3.14)/(((10.0^2-3.14)^2))`

`d=0.0120`

Put the value of
`\omega=6.28\ rad/s` in equation (I) and (II)

`A=((0.5)/(0.254))/(√((10.0^2-6.28)^2+0.628^2*6.28^2))`

`A=0.0209`

From equation (II)

`tan d=(0.628*6.28)/(((10.0^2-6.28)^2))`

`d=0.0257`

Put the value of
`\omega=9.42\ rad/s` in equation (I) and (II)

`A=((0.5)/(0.254))/(√((10.0^2-9.42)^2+0.628^2*9.42^2))`

`A=0.0217`

From equation (II)

`tan d=(0.628*9.42)/(((10.0^2-9.42)^2))`

`d=0.0413`

Hence, This is the required solution.