Question

A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x = 0.5 sin (pt+p/3). The acceleration (in m/s2) of the body at t = 1.0 s is approximately

a. 3.5
b. 49
c. 14
d. 43
e. 4.3

Answer 1

To find the acceleration at t = 1.0 s for a body oscillating with simple harmonic motion, differentiate the displacement equation twice with respect to time. Substitute the value of t into the acceleration equation to get the approximate value. None of the provided options are correct.

Step-by-step explanation:

The displacement of the body is given by the equation x = 0.5 sin (pt+p/3). To find the acceleration at t = 1.0 s, we need to differentiate the displacement equation twice with respect to time. The first derivative gives us the velocity, which is v(t) = 0.5p cos (pt+p/3), and the second derivative gives us the acceleration, which is a(t) = -0.5p^2 sin (pt+p/3).

Now, we can substitute t = 1.0 s into the acceleration equation and calculate the approximate value:

a(1.0) = -0.5p^2 sin (p + p/3) = -0.5p^2 sin (4p/3) = approximately -0.866p^2

Since we are not given the value of p, we cannot calculate the exact value of the acceleration. Therefore, none of the provided options (a, b, c, d, e) are correct.

Answer 2

I'll tell you how I look at this, although I may be missing something important.

Position = x(t) = 0.5 sin(pt + p/3)

Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)

At (t = 1.0),

x ' '(t) = -0.5 p² sin( 4/3 p )

In order to evaluate this, don't I still have to know what 'p' is ? ?

I don't think it can be evaluated with the information given in the question.