A mass m is attached to a spring of force constant 70 N/m and allowed to oscilla

Question

A mass m  is attached to a spring of force constant 70 N/m and allowed to oscillate. The figure (Figure 1) shows a graph of its velocity vx as a function of time t .

Part A

Find the period.

Enter your answer in seconds to three significant figures.

Part B

Find the frequency of this motion.

Express your answer in hertz to three significant figures.

Part C

Find the angular frequency of this motion.

Express your answer in radians per second to three significant figures.

Part D

What is the amplitude (in cm)?

Enter your answer in centimeters to two significant figures.

Part E

At what times does the mass reach the position x=±A?

Part F

Find the maximum acceleration of the mass.

Part G

Find the times at which the maximum acceleration occurs.

Explanation / Answer

a) by looking at the graph that the period is
T = (2.0 - 0.4) = 1.6 s

b) f = 1 / T = 0.625 Hz

c) = 2f = 3.927 rad/s

d)considering the max KE which is equal to the max PE.
max KE = 1/2mv2 = 1/2 * M * (20cm/s)2 = M * 200 cm²/s²
max PE = M * 200 cm²/s² = 1/2kx2
but since = sqrt(k/M), => k = M2,
M * 200 cm²/s² = 1/2(M2)x2 = 1/2M(3.927rad/s)2x2
x = 5.1 cm

e) x = ±A when v = 0, or when t = 0.4s, 1.2s

f) General form of velocity in SHM is
v(t) = Acos(t) = 5.1cm * 3.927rad/s * cos(3.927t) = 20cm/s * cos3.927t
Note that since A = Vmax, we have good confirmation on our previous calculations.
Then a(t) = A²sin(t) = 20cm/s * 3.927rad/s * sin(t) = 78.54cm/s² * sin(t)
so the max a = 78.54 cm/s²

g) at maximum acceleration sin(t) = 1

=> t = 1.571

=> t = 1.571/3.927 = 0.400051 sec

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