A particle executing simple harmonic motion of amplitude 5 cm and period 2 s. Find the speed of the particle at a point where its acceleration is half of its maximum value

x = 5sin(πt)

x" = -5π^2 sin(πt)

x" = 5/2 at two times. You can use the graph at

http://www.wolframalpha.com/input/?i=sin(t)+%3D+1%2F(2pi%5E2)

to see which value to use.

To find the speed of the particle at a point where its acceleration is half of its maximum value, we need to determine the maximum acceleration first.

In simple harmonic motion, the acceleration of a particle is given by the equation:

a = -ω²x

where:
a is the acceleration,
ω (omega) is the angular velocity (2π divided by the period T),
x is the displacement from the equilibrium position.

Given that the period T is 2 seconds, we can find ω:

ω = 2π / T
= 2π / 2
= π rad/s

Now, we know that the acceleration at any point is proportional to the displacement from the equilibrium position. The maximum acceleration (a_max) occurs at the extreme points of the motion, where the displacement is equal to the amplitude.

a_max = ω²A
= (π)²(5)
= 25π cm/s²

Next, we want to find the point where the acceleration is half of its maximum value.

a_half = 0.5 * a_max
= 0.5 * 25π
= 12.5π cm/s²

Let's denote the displacement at this point as x_half.

Now, we can rearrange the equation for acceleration to solve for displacement:

a = -ω²x

x = -a / ω²

Plugging in the values, we get:

x_half = - (12.5π) / (π)²
= -12.5 / π cm

To find the speed (v) of the particle at this point, we can use the relation between velocity and displacement in simple harmonic motion:

v = ω√(A² - x²)

Plugging in the values, we get:

v = π√(5² - (-12.5/π)²)
= π√(25 - (12.5/π)²)

Calculating this expression will give us the speed of the particle at the point where its acceleration is half of its maximum value.