Question

A body of mass 500 gm is attached to a spring and the system is driven by an external periodic force of amplitude 15N and frequency 0.796Hz. the spring extends by a length of 88mm under the given load. Calculate the amplitude of oscillation if the resistance coefficient of the medium is 5.05 kg/s​

Answer 1

Explanation:

Given A body of mass 500 gm is attached to a spring and the system is driven by an external periodic force of amplitude 15 N and frequency 0.796 Hz. the spring extends by a length of 88 mm under the given load. Calculate the amplitude of oscillation if the resistance coefficient of the medium is 5.05 kg/s

• Mass of the body m = 500 gm = 0.5 kg
• External periodic force F = 15 N
• Frequency f = 0.796 hz
• Length extended l = 88 m = 88 x 10^-3 m
• Resistance coefficient b = 5.05 kg / s
• Amplitude of oscillation A = ?
• Now we have A = F/m / √(ω^2 – p^2)^2 + (γp)^2
• We have γ = b/m
•                  = 5.05 / 0.5
•                = 10.1 s^-1
• Angular frequency of applied force p = 2πf
•                                                                = 2 x 3.14 x 0.796
•                                                                  = 4.9
•                                                                 = 5 rad / s
• Force constant k = F / l  
•                             = mg / l
•                            = 0.5 x 9.8 / 88 x 10^-3
•                           = 55.68 N/m
• Frequency ω = √k / m
•                       = √55.68 / 0.5
•                      = √111.36
•                       = 10.55 rad / s              
• Substituting all the above values in the equation we get
•                               A = F/m / √(ω^2 – p^2)^2 + (γp)^2
•                                   = 15 / 0.5 / √[(10.55)^2 – 5^2]^2 + (10.1 x 5)^2
•                                 = 30 / √(111.3025 – 25)^2 + (50.5)^2
•                                  = 30 / √7448.12 + 2550.25)
•                                = 30 / 99.991
•                             A = 0.3 m
• Therefore the amplitude of oscillation A = 0.3 m

Reference link will be

https://brainly.in/question/3118996

Answer 2

Given that,

Mass of body m= 500 gm

Force F= 15 N

Frequency f= 0.796 Hz

Length l= 88 mm

Resistance coefficient b= 5.05 kg/s

We need to calculate the angular frequency

Using formula of angular frequency of applied force

$p=2\pi f$

Put the value into the formula

$p=2\pi\times0.796$

$p=5.00\ rad/s$

We need to calculate the force constant

Using formula of force constant

$k=\dfrac{F}{l}$

Put the value into formula

$k=\dfrac{0.5*9.8}{88\times10^{-3}}$

$k=55.68\ N/m$

We need to calculate the frequency

Using formula of frequency

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

Put the value into the formula

$\omega=\sqrt{\dfrac{55.68}{0.5}}$

$\omega=10.55\ rad/s$

We need to calculate the value of γ

Using formula of γ

$\gamma=\dfrac{b}{m}$

Put the value into the formula

$\gamma=\dfrac{5.05}{0.5}$

$\gamma=10.1\ s^{-1}$

We need to calculate the amplitude of oscillation

Using formula of amplitude

$A=\dfrac{\dfrac{F}{m}}{\sqrt{(\omega^2-p^2)^2+(\gamma p))^2}}$

Put the value into the formula

$A=\dfrac{\dfrac{15}{0.5}}{\sqrt{(10.55^2-5^2)^2+(10.1\times5)^2}}$

$A=0.3\ m$

Hence, The amplitude of oscillation is 0.3 m