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

Answer:

A. 2.249 hz

B. 0.45 s

C. 0.424 m/s

D. 66⁰

E. 6 m/s^2

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

$f=\frac{1}{2\pi } \sqrt \frac{k}{m}$

$f=\frac{1}{2\pi } \sqrt \frac{150}{0.75}$

f = 2.249 hz

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

$period (T) = \frac{1}{frequency}$

$T = \frac{1}{2.249} (s)$

T = 0.45 s

Step 4: calculate the maximum velocity,$V_{max}$

$V_{max} = A\sqrt{\frac{k}{m} }$

A is the amplitude of the oscilation

$V_{max} = 0.03\sqrt{\frac{150}{0.75} }$

$V_{max} = 0.424(\frac{m}{s})$

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

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{150}{0.75} }$

$\omega= 14.142$

$\phi = phase angle$

Applying displacement equation in S.H.M

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

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

$cos(6.364+\phi) = \frac{0.009}{0.03}$

$cos(6.364+\phi) = 0.3$

$(6.364+\phi) = cos^{-1}(0.3)$

$(6.364+\phi)= 72.5⁰$

$6.364+\phi =72.5⁰$

$\phi$ =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 $(\frac{m}{s^{2} })$

$a_{max}$ ≅ 6 $(\frac{m}{s^{2} })$

Answer 2

Answer:

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) $a_{max}=6.0041 [tex]\frac{m}{s^2}$

Explanation:

a)

Natural frequency

  $\omega _n=\sqrt {\dfrac{K}{m}}$

$\omega _n=\sqrt {\dfrac{150}{0.75}}$

$\omega _n$=14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

$=\dfrac{2π}{\omega _n}$

T=$\frac{1}{f}$

 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=$\frac{dx}{dt}$

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)

$V_{max}=\omega_n X_{max}$

$V_{max}=14.14\times 0.03003$=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 $\frac{m}{s^2}$