The position of a particle undergoing simple harmonic motion is given by x(t)=30

Question

The position of a particle undergoing simple harmonic motion is given by x(t)=30cos(4t), where x is in millimeters and t is in seconds. A) Amp = 30 mm B) Frequency = 2 Hz C) Period = 0.50 s D) What is the first instant at which the particle is at x = 0? = 0.13 s E) What is the second instant at which the particle is at x = 0?= 0.38 s F) What is the third instant at which the particle is at x = 0 = 0.63 s.

G) Determine the x component of position of the particle at t = 1.5 s.

H) Determine the x component of velocity of the particle at t = 1.5 s.

I) Determine the x component of acceleration of the particle at t = 1.5 s.

Explanation / Answer

given,

equation of shm x(t)=30cos(4t)

equation od shm is A * cos(wt)

where,

A = amplitude, w = frequency

so, amplitude for this motion = 30 mm   (Answer a)

w = 2*pi*f = 4pi

frequency f for this motion = 2 Hz (Answer b)

Time period for this motion = 1/frequency

Time period for this motion = 1/2 = 0.5 sec (Answer c)

For x = 0

x(t) = 30 * cos(4 **t)

30 * cos(4**t) = 0

cos(4 **t) = cos(/2)

(4 **t) = (2n-1)(/2)

For x to be 0, for the first time n =1

t = (/2)/ (4*)

t = 1/8 = 0.125 s (Answer d)

For x to be 0, for the first time n =2

t = (3/2)/ (4*)

t = 3/8 = 0.37 s (Answer e)

For x to be 0, for the first time n =3

t = (5/2)/ (4*)

t = 5/8 = 0.625 s (Answer f)

x(t)=30cos(4t)

x(1.5) = 30* Cos(4**1.5) = 28.4 mm ((Answer g)

Velocity is

v(t) at t=1.5 s is v(1.5) = - 30*4* Sin( 4**1.5) = -121.8 mm/s = -0.122 m/s (Answer h)

Acceleration is

a(t) = -30*(4)^2 Cos(4*1.5) = -4483.3 mm/s2 = -4.483 m/s2 (Answer i)

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