A 3.0 kg frictionless block is attached to an ideal spring of force constant 300

Question

A 3.0 kg frictionless block is attached to an ideal spring of force constant 300 N/m. The block is then set into oscillation with an amplitude of 30.0 cm, and in such a way that the mass passes through the equilibrium point (Xo) on its way towards negative displacement at time t = 0 s. For this block's oscillation, k = 300 N/m 3.0 kg -30.0 cmXo-0 cm 30.0 cm Figure 1 i) identify the equation that describes this oscillation as a function of time, calculate the time needed for the block to go from x=-15.0 cm right up to the x = +25.0 cm and ii find the time, t, when it passes the x10.0 cm point for the 7th time.

Explanation / Answer

part i.

mass of the block=m=3 kg

spring constant=k=300 N/m

then angular frequency=w=sqrt(k/m)=10 rad/s

oscillation amplitude=A=30 cm=0.3 m

as oscillation a sinusoidal movement,

position x at any time t is given as

x(t)=-A*sin(w*t)=-30*sin(10*t) cm

negative sign is because of the fact that it is moving to the -ve direction at t=0

part ii.

let the block is at x=-15 cm at t=T1 and at x=25 cm at t=T2

then -15=-30*sin(10*T1)

==>T1=0.05235 seconds

and 25=-30*sin(10*T2)

==>T2=0.41267 seconds

then time needed=T2-T1=0.36032 seconds

part iii.

time period=2*pi/w=pi/5 second

let at time t , it will reach x=10 cm for the first time

then 10=-30*sin(10*T)

==>T=0.348 seconds

it will cross x=10 cm at T+time period=0.9763 seconds

and so on

so it will pass for the 7th time at T+6*time period=4.118 seconds

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