Answer to Question #188471 in Mechanics | Relativity for Fadiora damilola

Question #188471

1. Express the displacement of velocity and acceleration as a function of time


2. Express velocity and acceleration as a function of time


3. A mass of 100g is connected is a light spring of force constant 10Nm and is free to oxillate on a horizontal frictionless surface. If the mass is displaced 8cm from equilibrium and release from the rest. Find: i. The period of it's motion ii. The maximum speed of the mass iii. The maximum acceleration of the mass


1
Expert's answer
2021-05-06T17:22:04-0400

Solution :-

Displacement x=asin(wt+ϕ)x=asin(wt+\phi)

(a) Velocity in displace ment form

v=dxdt=awv=\frac{dx}{dt}=aw

Acceleration in displacement form


a=d2xdt2=w2xa=\frac{d^2x}{dt^2}=w^2x

(b) Acceleration in term of velocity

a=dvdt=w2xa=\frac{dv }{dt}=w^2x


(c) we know that

Kx =mg

We know that

T=2πmkT=2\pi \sqrt\frac {m}{k}

Put value

TT =2πmk=2π100×10310=0.628sec=2\pi \sqrt\frac {m}{k}=2\pi \sqrt\frac {100\times10^{-3}}{10}=0.628sec

T=0.628sec

Maximum speed 0.8m/sec

Maximum speed(v)=ωx=2πn(x)\omega x=2\pi n(x)

v=2πTxv=\frac{2\pi}{T}x

v=2π0.628×.08=0.8m/secv=\frac{2\pi}{0.628}\times.08=0.8m/sec

Put values

V=0.8m/sec

Maximum acceleration

a=ω2xa=\omega^2x

Put value a=(2πn)2xa=(2\pi n)^2x

Put value

a=(2πT)2xa=(\frac{2π}{T})^2x

a=(2×3.140.628)2×0.08a=(\frac{2\times3.14}{0.628})^2\times0.08

a=8m/sec2a=8m/sec^2

a=9.80m/sec2a=9.80m/sec^2


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment