A block attached to a spring is displaced from equilibrium to the position x=+2.8m and released. The period is 1.6s . At what positions and times during the first complete cycle do the following conditions occur.

a) |v|=0.5vmax, Where vmax is the maximum speed?

First position:
First time:


Second position:
Second time:


Third position:
Third time:


Fourth position:
Fourth time:
s.

b) |a|=0.5amax , Where amax is the maximum magnitude of the acceleration?

First position:
First time:


Second position:
Second time:


Third position:
Third time:


Fourth position:
Fourth time:

Please helpp im desperate!!

To find the positions and times during the first complete cycle when certain conditions occur, we need to determine the mathematical expressions for velocity and acceleration of the block attached to the spring.

Let's start by finding the expressions for velocity and acceleration in simple harmonic motion where T represents the period and x represents the displacement from equilibrium:

Velocity (v) = ωx
Acceleration (a) = -ω²x

Here, ω represents the angular frequency and is calculated as ω = 2π / T.

Given the period T = 1.6s, we can calculate ω as follows:

ω = 2π / T = (2π) / (1.6s) = π / 0.8 s⁻¹ = 1.25π s⁻¹

Now, let's find the maximum speed (vmax) and maximum magnitude of acceleration (amax) using the expressions mentioned above:

vmax = ωA
amax = ω²A

In simple harmonic motion, the amplitude (A) represents the maximum displacement from equilibrium. In this case, it's given that the displacement from equilibrium is x = +2.8m.

Substituting the values, we can find vmax and amax:

vmax = (1.25π s⁻¹) × (+2.8m) ≈ 3.493 m/s
amax = (1.25π s⁻¹)² × (+2.8m) ≈ 10.864 m/s²

Now, we can determine the positions and times during the first complete cycle when |v| = 0.5vmax and |a| = 0.5amax.

a) |v| = 0.5vmax:

To find the positions, we know that |v| = vmax when the block is at the extreme positions of the motion. Therefore, the desired positions occur when |v| = 0.5vmax.

At extreme positions, velocity is maximum, and the block momentarily stops before changing direction. This occurs when the velocity is zero:

First position: When |v| = 0.5vmax, the block is at the extreme position. Since |v| = 0.5vmax = 0.5 × 3.493 m/s ≈ 1.747 m/s, the first position occurs when x = ±1.747 m.

To find the time for the first position, we need to consider the time taken for the block to reach that position from equilibrium. Since simple harmonic motion is symmetrical, the time taken to reach the first position is half the period:

First time: t = T / 2 = 1.6s / 2 = 0.8s.

Next, we can find the second position and time:

Second position: When |v| = 0.5vmax, the block is again at the extreme position. Therefore, the second position occurs when x = ±1.747 m (same as the first position).

Second time: Since the block is at the extreme position again, the time taken would be equivalent to the time for the first position, which is 0.8s.

Now, let's move on to find the third and fourth positions:

Third position: When |v| = 0.5vmax, the block is at the equilibrium position, where the velocity is maximum. Therefore, the third position occurs when x = 0m.

To find the time for the third position, we need to consider the time taken for the block to reach equilibrium from the extreme position. Since simple harmonic motion is symmetrical, the time taken to reach the equilibrium position is half the period again:

Third time: t = T / 2 = 0.8s.

Finally, the fourth position occurs at the next extreme position:

Fourth position: When |v| = 0.5vmax, the block is at the extreme position once more, where x = ±1.747m.

Fourth time: Since the block is at the extreme position, the time taken would be equivalent to the time for the first position, which is 0.8s.

Therefore, the answers are:

a) |v| = 0.5vmax:
First position: x = ±1.747m
First time: t = 0.8s

Second position: x = ±1.747m
Second time: t = 0.8s

Third position: x = 0m
Third time: t = 0.8s

Fourth position: x = ±1.747m
Fourth time: t = 0.8s

b) |a| = 0.5amax:

Similar to the above process, we can determine the positions and times during the first complete cycle when |a| = 0.5amax.

Please let me know if you would like to proceed with part b) or need any further assistance.

cosine 0 =1 so

x = A cos 2 pi t/T = 0.28 cos 2 pi t/T
but T = 1.6
sso
x = 0.28 cos 3.93 t
note ---- 3.93 t is in radians not degrees
now
dx/dt = v = - 0.28 * 3.93 sin 3.93 t = -1.10 sin 3.93 t
and
a = d/dt v = -1.10* 3.93 * cos 3.93 t = -4.32 cos 3.93 t
==================== now plug and chug
for example |v| is max first when 3.93 t = pi/2 (or 90 degrees|
t = 3.14159 / (3.93*2) = 0.40 seconds
and velocity v is - 1.10 there)

if you want to know for exmple when v = 1/2 v max it its when the sin is 1/2

that is when the angle is 30 degrees which is pi/6 radians