Given the equation for simple harmonic method d=1.3cos(3pi/4 t) answer the following questions:

a)Find the maximum displacement.
1.i got 1.3

b)Find the frequency and length of the pendulum

c)Calculate the value of d when t=3
i got .919

d)Determine the least positive value of t for which d=0.
i got 2/3

b) To find the frequency and length of the pendulum, we need to use the formula for simple harmonic motion, which is:

d = A * cos(ωt)

where:
- d is the displacement
- A is the amplitude (maximum displacement)
- ω is the angular frequency
- t is the time

Comparing the given equation with the formula, we have A = 1.3 and ω = 3π/4.

The frequency (f) and angular frequency (ω) are related by the equation ω = 2πf. Therefore, we can calculate the frequency using:

ω = 3π/4 = 2πf

Solving for f:

3π/4 = 2πf
f = (3/4) / (2π)
f = 3/8π

The length of the pendulum (L) can be calculated using the relationship between the angular frequency and the length of the pendulum:

ω = √(g/L)

Where g is the acceleration due to gravity. Rearranging the equation, we get:

L = g / ω^2

Given that g ≈ 9.8 m/s^2, we can substitute the values:

L = 9.8 / (3π/4)^2
L = 9.8 / (9π^2/16)
L = 157.08 / π^2

So the frequency is 3/8π and the length of the pendulum is approximately 50.1 units.

c) To calculate the value of d when t = 3, we can substitute t = 3 into the equation:

d = 1.3 * cos(3π/4 * 3)
d = 1.3 * cos(9π/4)
d = 1.3 * cos(2.25π)
d ≈ 0.919

So when t = 3, the value of d is approximately 0.919.

d) To determine the least positive value of t for which d = 0, we can set d = 0 in the equation and solve for t:

0 = 1.3 * cos(3π/4 * t)

To find the least positive value of t, we need to find the first zero-crossing. The period of the cosine function is 2π, so we need to find the smallest value of t that satisfies this equation within 0 to 2π.

Setting the equation equal to zero:

cos(3π/4 * t) = 0

Solving for t:

3π/4 * t = π/2 + nπ, for n = 0,1,2,...

t = (2/3) + (8/3)n, for n = 0,1,2,...

The smallest positive value of t will occur when n = 0:

t ≈ 2/3

So the least positive value of t for which d = 0 is approximately 2/3.

a) To find the maximum displacement, you need to look at the coefficient of the cosine function. In this case, the coefficient is 1.3, which means that the maximum displacement is 1.3 units.

b) The equation for simple harmonic motion can also be written as d = Acos(ωt), where A is the amplitude and ω is the angular frequency. Comparing this equation with the given equation, we can infer that A = 1.3.

For the frequency, we know that ω = 2πf, where f is the frequency. In the given equation, ω = (3π/4). Substituting this value into the equation ω = 2πf, we can solve for f:

(3π/4) = 2πf
f = (3π/4) / (2π)
f = 3/8

So the frequency of the pendulum is 3/8 cycles per unit of time.

To find the length of the pendulum, we can use the formula T = 2π√(L/g), where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.

The period T is the reciprocal of the frequency f, so T = 1/f = 8/3. Substituting this into the formula and solving for L:

8/3 = 2π√(L/g)
4/3 = π√(L/g)
16/9 = π^2 (L/g)
L/g = 16/(9π^2)
L = (16/(9π^2)) * g

So the length of the pendulum is (16/(9π^2)) times the acceleration due to gravity.

c) To calculate the value of d when t = 3, substitute t = 3 into the given equation:

d = 1.3cos(3π/4 * 3) = 1.3cos(9π/4) ≈ 0.919

So the value of d when t = 3 is approximately 0.919.

d) To determine the least positive value of t for which d = 0, set d equal to 0 and solve for t:

0 = 1.3cos(3π/4 * t)
cos(3π/4 * t) = 0

The cosine function is zero at multiples of π/2. So, we need to find the smallest positive value of t that makes 3π/4 * t a multiple of π/2.

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

So the least positive value of t for which d = 0 is 2/3.

(b)The period is

T=2π/(3π/4) = 8/3 s, so the frequency is 3/8 Hz

Now, use T=2π√(L/g) to find L

(d) since cos(π/2)=0, you want

3π/4 t = π/2
t = 3π/4 * 2/π = 3/2

for L i got 3.85