Question

se/ViewProblem?unititemID=4124598&enrollmentID=389313 A block with mass m 7.2 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x 0.27 m. While at this equilibrium position, the mass is then given an initial push downward at v-4.8 m/s. The block oscillates on the spring without friction. 1) What is the spring constant of the spring? 261.3 3 Your submissions: 261.31 Computed value: 261.3 Feedback: Correct! Submitted: Thursday, October 3 at 12:26 AM 2) What is the oscillation frequency? 0.958 Hz Submit Your submissions: 0.958 Computed value: 0.958 Feedback: Correct! Submitted: Thursday, October 3 at 12:28 AM

I need only help with 3,4,and 5

Answer 1

The angular frequency of the oscillation in the spring-mass system is given by

一

Putting the values in SI units

261.33 – 6,0246 rad/s

The equation of motion of SHM is given by

y= Asin (wt)

Let us now calculate the amplitude A of the oscillations using energy conservations.

At the equilibrium point, the energy is

E-?mo_ 7.2 x 4.82 = 82.944 J

And at one end the energy is given by

==kA

Conservation of energy gives us

ᎭᎭᏮ 8 = zᏙ1-

2 x 82.944 A= 261.33 = 0.7967 m

So, the equation of oscillation becomes

y = 0.7967 sin (6.0246t)

The velocity as a function of time is

v(t) = y = 0.7967 x 6.0246 x cos (6.0246t) = 4.8 cos(6.0246t)

And the acceleration as a function of time is

a(t) = u(t) = -4.8 x 6.0246 sin(6.0246t) = -28.918 sin(6.0246t)

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PART 3:

Speed of the block at t = 0.43 s is

$v(0.43)=4.8\cos(6.0246\times 0.43)=4.795\ m/s$

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PART 4:

The magnitude of maximum acceleration is

$a(t)=\left |-28.918\times (1) \right |=28.918\ m/s^2\ \ \ \ \ \ \because \sin(\theta)_{max}=1$

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PART 5:

At t = 0.43 s, the magnitude of the acceleration of the block is

a(0.43) = -28.918 sin(6.0246 x 0.43) = -1.307 m/s

So, the magnitude of the force on the block is

$F=\left |ma \right |=\left | 7.2\times (-1.307) \right |=9.41\ N$