Question

The diagram above shows the motion of a 2.00−kg mass on a horizontalspring. Draw the reference circle. Find the phase

constant. Write down the equation of the displacement as a function of time. What is the spring constant? What is the total

energy? What is the maximum speed? What is the maximum acceleration? When exactly will the mass be at equilibrium and

moving to the right? When exactly will the mass be at point C?

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Answer 1

$\huge\fcolorbox{black}{aqua}{Solution:-}$

✏️ Answer:-

☛ At 0.277s the mass will be exactly at point C.

✏️ Explanation:-

☛ Examining the given graph,

We see that the largest displacement is 10 cm,

⇒A = 0.10 m.

We also see that the motion repeats every

0.2 seconds, so T = 0.2 s. Now, as know, angular frequency is related to the period by ω = 2π/T,

⇒ω = 10π.

At t = 0, we can read from the side of the graph that x = - ­7.5 cm = ­-0.075 m.

[Note- The negative x position puts the object at

either quadrant II or III of the reference circle. To decide which quadrant is correct, note that as t increases the curve goes

through zero (equilibrium), labelled point A in the diagram, and it is moving from negative to positive position so it is moving to the right.]

★ This is the behaviour of the particle in quadrant III, so the reference circle looks like the one provided in the 2nd attachment. (Kindly have a glimpse at it! (: )

We can take the general equation, x = Acos(ωt + ϕ0) and substitute in the t = 0 values to find ϕ0

-0.075 = 0.10cos (0 + ϕ0),

which reduces to

-­0.75 = cos(ϕ0) ,

⇒ϕ0 = cos^-1(-0.75).

Now,

Solving yields 138.6° (2.419 rad) or 221.4° (3.864 rad). The second answer is in the third quadrant and is thus the correct phase constant.

Thus,

The equation of the displacement is

x = 0.10cos(10πt + 3.864) ,

An examination of our relationships for SHM indicates that

K = ω²m, E = ½KA²,

Vmax = ωA, and amax = ω²A, so

K = (10π /s)²(2kg) = 1974 N/m.

E = ½(1974 N/m)(0.1m)² = 9.870 J.

V ᵐᵃˣ = (10π/s)(0.1m) = π m/s.

aᵐᵃˣ = (10 π/s)²(0.1m) = 9870 m/s².

Now,

★ To find when the object is moving to the right at equilibrium (labelled A in the given x­t graph), note that this is point 3 on the reference circle given above. So the object needs to rotate to 270° from 221.4°. Since rotating 360° takes one period which here is 0.2 s, tA = 48.6/360 × 0.2s = 0.027 s.

Point C in the diagram occurs at point 1 on the reference circle but note that point C is more than one period T from t = 0.

We need to calculate the time to rotate to point 1 but then add one period. The angle we need to rotate through is 360° − 221.4° = 138.6°.

⇒tC = 138.6/360 × 0.2 s + 0.2 s = 0.277 s.

∴ At 0.277s the mass will be exactly at point C.

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I hope this helps! :)