Question 4
A 0.9 kg block attached to a spring of force constant 2.1 N/m oscillates with an amplitude of 9 cm. A)Find the maximum speed of the block. Answer in units of m/s
B)Find the speed of the block when it is 4.5 cm from the equilibrium position.
Answer in units of m/s
C)Find its acceleration at 4.5 cm from the equilibrium position. Answer in units of m/s^2
D)Find the time it takes the block to move from x = 0 to x = 4.5 cm. Answer in units of s
a = .09 meter
w = 2 pi f = sqrt (k/m) = sqrt(2.1/.9) = 1.53 radians/s
x = a sin w t
x = .09 sin 1.53 t
v = .09*1.53 cos 1.53 t = .1377 cos .153 t
That is max when cos = 1 so part a is .1377 m/s
a = -.09*1.53^2 sin wt = - 1.53^2 x
when x = .045
sin 1.53 t = .045/.09 = .5
so 1.53 t = pi/6 radians (30 degrees)
then t = .342 seconds (answer to part d by the way)
cos 30 degrees = .866
so
v = .1377*.866 = .119 m/s (part b)
a = -1.53^2*.09 * sin 30 = .105 m/s^2
Wow thank you. I was really lost.
this is a huge help! thanks again!
In order to answer these questions, we need to use the equations of motion for simple harmonic motion (SHM).
The equations we will be using are:
1. Maximum velocity: vmax = Aω, where A is the amplitude and ω is the angular frequency given by ω = √(k/m), where k is the force constant and m is the mass of the block.
2. Velocity at any position: v = ω√(A^2 - x^2), where x is the displacement from the equilibrium position.
3. Acceleration at any position: a = -ω^2x, where x is the displacement from the equilibrium position.
4. Time period: T = 2π/ω.
Now, let's solve each part of the question:
A) To find the maximum speed of the block, we can use the equation vmax = Aω.
Given:
Amplitude (A) = 9 cm = 0.09 m
Force constant (k) = 2.1 N/m
Mass (m) = 0.9 kg
First, we need to find the angular frequency:
ω = √(k/m) = √(2.1/0.9) = 1.4 rad/s
Now, we can find the maximum velocity:
vmax = Aω = 0.09(1.4) = 0.126 m/s
So, the maximum speed of the block is 0.126 m/s.
B) To find the speed of the block when it is 4.5 cm from the equilibrium position, we can use the equation v = ω√(A^2 - x^2).
Given:
Amplitude (A) = 9 cm = 0.09 m
Displacement (x) = 4.5 cm = 0.045 m
We already calculated the angular frequency:
ω = 1.4 rad/s
Now, we can find the speed at this position:
v = ω√(A^2 - x^2) = 1.4√(0.09^2 - 0.045^2) = 1.37 m/s
So, the speed of the block when it is 4.5 cm from the equilibrium position is 1.37 m/s.
C) To find the acceleration at 4.5 cm from the equilibrium position, we can use the equation a = -ω^2x.
Given:
Displacement (x) = 4.5 cm = 0.045 m
We already calculated the angular frequency:
ω = 1.4 rad/s
Now, we can find the acceleration at this position:
a = -ω^2x = -(1.4)^2(0.045) = -0.0876 m/s^2
So, the acceleration at 4.5 cm from the equilibrium position is -0.0876 m/s^2.
D) To find the time it takes the block to move from x = 0 to x = 4.5 cm, we need to find the time period.
Given:
Amplitude (A) = 9 cm = 0.09 m
Force constant (k) = 2.1 N/m
Mass (m) = 0.9 kg
First, we need to find the angular frequency (ω) as we did in part A.
Now, we can find the time period:
T = 2π/ω = 2π/1.4 = 4.51 s
So, it takes the block 4.51 seconds to move from x = 0 to x = 4.5 cm.