Question

A block of mass 0.75 kg is suspended from a spring having a stiffness of 150 N/m. The block is displaced downwards from its equilibrium position through a distance of 3 cm with an upward velocity of 2 cm/sec. Determine: a)- The natural frequency b)-The period of oscillation c)-The maximum velocity d)-The phase angle e)-The maximum acceleration

Answer 1

A. 2.249 hz

B. 0.45 s

C. 0.424 m/s

D. 66⁰

E. 6 m/s^2

Step-by-step explanation:

Step 1: identify the given parameters

mass of the block (m)= 0.75kg

stiffness constant (k) = 150N/m

Amplitude (A) = 3cm = 0.03m

upward velocity (v) = 2cm/s

Step 2: calculate the natural frequency (F)by applying relevant formula in S.H.M

`f=(1)/(2\pi ) \sqrt (k)/(m)`

`f=(1)/(2\pi ) \sqrt (150)/(0.75)`

f = 2.249 hz

Step 3: calculate the period of the oscillation (T)

`period (T) = (1)/(frequency)`

`T = (1)/(2.249) (s)`

T = 0.45 s

Step 4: calculate the maximum velocity,
`V_(max)`

`V_(max) = A\sqrt{(k)/(m) }`

A is the amplitude of the oscilation

`V_(max) = 0.03\sqrt{(150)/(0.75) }`

`V_(max) = 0.424((m)/(s))`

Step 5: calculate the phase angle, by applying equation in S.H.M

`X = Acos(\omega{t} +\phi)`

where X is the displacement; calculated below

Displacement = upward velocity X period of oscillation

`displacement (X) = vt (cm)`

X = (2cm/s) X (0.45 s)

X = 0.9 cm = 0.009m

where
`\omega` is omega; calculated below

`\omega=\sqrt{(k)/(m) }`

`\omega=\sqrt{(150)/(0.75) }`

`\omega= 14.142`

`\phi = phase angle`

Applying displacement equation in S.H.M

`X = Acos(\omega{t}+\phi)`

`0.009 = 0.03cos(14.142 X 0.45+\phi)`

`cos(6.364+\phi) = (0.009)/(0.03)`

`cos(6.364+\phi) = 0.3`

`(6.364+\phi) = cos^(-1)(0.3)`

`(6.364+\phi)= 72.5⁰`

`6.364+\phi =72.5⁰`

`\phi` =72.5 -6.364

`\phi` =66.1⁰

Phase angle,
`\phi` ≅66⁰

Step 6: calculate the maximum acceleration,
`a_(max)`

`a_(max) = \omega^(2)A`

`a_(max)` = 14.142 X 14.142 X 0.03

`a_(max)` = 5.999
`((m)/(s^(2) ))`

`a_(max)` ≅ 6
`((m)/(s^(2) ))`

Answer 2

a)f=2.25 Hz

b)Time period T=.144 s

c)tex]V_{max}[/tex]=0.42 m/s

d)Phase angle Ф=87.3°

e)
`a_(max)=6.0041 [tex](m)/(s^2)`

Step-by-step explanation:

a)

Natural frequency

`\omega _n=\sqrt {(K)/(m)}`

`\omega _n=\sqrt {(150)/(0.75)}`

`\omega _n`=14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

`=(2π)/(\omega _n)`

T=
`(1)/(f)`

Time period T=.144 s

c)Displacement equation

`x=Acos\omega _nt+Bsin\omega _nt`

Boundary condition

t=o,x=0.03 m

t=0,v=.02m/s , V=
`(dx)/(dt)`

Now by using these above conditions

A=0.03,B=0.0014

x=0.03 cos14.14 t+0.0014 sin14.14 t

⇒x=0.03003sin(14.14t+87.3)

`V_(max)=\omega_n X_(max)`

`V_(max)=14.14* 0.03003`=0.42 m/s

d)

Phase angle Ф=87.3°

e)

Maximum acceleration

`a_(max)=(\omega _n )^2X_(max)`

`a_(max)=(14.14)^20.03003`=6.0041
`(m)/(s^2)`