Frequency and amplitude of a particle in simple harmonic motion a particle moves in simple harmonic motion. knowing that the maximum velocity is 200 mm/s and the maximum acceleration is 13 m/s , determine the amplitude and frequency of the motion. amplitude of the motion xm 1mm. frequency of the motion勿 hz.

amplitude, a = 3.076 mm

frequency, f = 10.35 Hz

Explanation:

Vmax = 200 mm /s = 0.2 m/s

Amax = 13 m/s

Let f be the frequency and a be the amplitude.

Use the formula of maximum velocity.

Vmax = ω a

0.2 = ω a     (1)

Use the formula of maximum acceleration.

Amax = ω^2 x a

13 = ω^2 a     (2)

Divide equation (2) by (1)

ω = 13 / 0.2 = 65 rad/s

Put in equation (1)

0.2 = 65 x a

a = 3.076 x 10^-3 m

a = 3.076 mm

Let f be the frequency

ω = 2 π f

f = 65 / (2 x 3.14) = 10.35 Hz  

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

grav forces affect all objects magnetic forces affect magnetic objects and electric currents