A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum accelera

Question

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, and an unknown phase constant ?.
(a) What is the period of the motion?
1 s

(b) What is the maximum speed of the particle?
2 m/s

(c) What is the total mechanical energy of the oscillator?
3 J

(d) What is the magnitude of the force on the particle when the particle is at its maximum displacement?
4 N

(e) What is this force when the particle is at half its maximum displacement?
5 N

Explanation / Answer

Mass particle m = 10 g
                         = 0.01 kg
Amplitude A = 2.0 mm
                    = 2 x 10 -3 m
Maximum acceleration a = 2.0 x 10 3 m/s 2
We know a = A 2
From this angular frequency = [ a/A]
                                              = 1000 rad / s
(a) The period of the motion T = 2 /
                                               = 6.283 x 10 -3 s
(b) The maximum speed of the particle V = A
                                                               = 2 m / s
(c) The total mechanical energy of the oscillator E = ( 1/ 2) mV 2
                                                                           = 0.02 J
(d) The magnitude of the force on the particle when the particle is at its maximum displacement
   F = kA
Where k = force constant
                = m 2
                = 10000 N / m
So, F = 20 N
(e) Force when the particle is at half its maximum displacement F ' = kx
Where x = displacment
               = A/ 2
So, F ' = 10 N

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