A 2kg object is attached to a spring and undergoes simple harmonic motion. At t=0 the object starts from rest, 10 cm from the equilibrium position. If the force constant of the spring k is equal to 75 N/m, calculate

a. the max speed
b. the max acceleration of the object. Fine the velocity of the object at t=5s

10 cm = 0.1 m

x = .1 cos 2 pi t/T

v = -(2 pi/T) .1 sin 2 pi t/T

a = - (2 pi/T)^2 .1 cos 2 pi t/T = -(2 pi/T)^2 x

F = - k x

F = m a = - 2 (2 pi/T)^2 x
k= 8 pi^2/T^2 = 75

T ^2 = 8 pi^2/75
T = 1.03 s
2pi /T = 6.12
(2 pi/T)^2 = 37.5

so
if v = -(2 pi/T) .1 sin 2 pi t/T
then v max = (2 pi/T) .1
=.612
and
if a = - (2 pi/T)^2 .1 cos 2 pi t/T
then a max = .1(2 pi/T)^2
=3.75

To solve this problem, we can use the equations of simple harmonic motion (SHM).

The equation for the maximum speed of an object undergoing simple harmonic motion is given by:

v_max = A * ω

where v_max is the maximum speed, A is the amplitude of the motion, and ω is the angular frequency.

a. The amplitude of the motion can be found using the given information:
Amplitude, A = 10 cm = 0.1 m

The angular frequency, ω, can be calculated using the formula:
ω = √(k / m)

where k is the force constant of the spring and m is the mass of the object.

Given:
k = 75 N/m
m = 2 kg

Substituting the given values in the formula, we get:
ω = √(75 N/m / 2 kg) = √(37.5 N/kg) ≈ 6.1237 rad/s

Now, we can calculate the maximum speed using the formula:
v_max = A * ω = 0.1 m * 6.1237 rad/s ≈ 0.6124 m/s

Therefore, the maximum speed of the object is approximately 0.6124 m/s.

b. The maximum acceleration, a_max, of the object undergoing simple harmonic motion is given by:
a_max = A * ω^2

Substituting the given values, we have:
a_max = 0.1 m * (6.1237 rad/s)^2 ≈ 0.3749 m/s^2

Therefore, the maximum acceleration of the object is approximately 0.3749 m/s^2.

To find the velocity of the object at t = 5 s, we can use the equation of motion for simple harmonic motion:

v = A * ω * cos(ωt + φ)

where v is the velocity, A is the amplitude of the motion, ω is the angular frequency, t is the time, and φ is the phase angle.

At t = 0, the object starts from rest, so the phase angle φ is zero.

Substituting the given values, we can calculate the velocity at t = 5 s as follows:

v = 0.1 m * 6.1237 rad/s * cos(6.1237 rad/s * 5 s + 0) ≈ 0.0612 m/s

Therefore, the velocity of the object at t = 5 s is approximately 0.0612 m/s.

To calculate the maximum speed and maximum acceleration of the object undergoing simple harmonic motion, we need to use the formulas for these quantities.

a. Maximum Speed:
The maximum speed of an object undergoing simple harmonic motion can be found using the formula:
v_max = Aω

First, we need to find the angular frequency (ω) of the object. The angular frequency can be calculated using the formula:
ω = sqrt(k/m)

where k is the force constant of the spring and m is the mass of the object. In this case, k = 75 N/m and m = 2 kg.

ω = sqrt(75 N/m / 2 kg)

After calculating ω, we can find the maximum speed (v_max) by multiplying it by the amplitude of the motion (A). In this case, the amplitude is given as 10 cm, which is equal to 0.1 m.

v_max = 0.1 m * ω

b. Maximum Acceleration:
The maximum acceleration of an object undergoing simple harmonic motion can be found using the formula:
a_max = Aω^2

After calculating ω, we can find the maximum acceleration (a_max) by multiplying it by the square of the angular frequency.

a_max = 0.1 m * ω^2

To find the velocity of the object at t = 5s, we can use the formula for velocity in simple harmonic motion:
v = Aω * cos(ωt + φ)

where t is the time elapsed and φ is the phase constant. Since the object starts from rest, the initial phase is 0.

v = 0.1 m * ω * cos(ωt)

To find the velocity at t = 5s, substitute t = 5s into the equation:
v = 0.1 m * ω * cos(ω * 5s)

Now, let's calculate these values.