A 100-g block hangs from a spring with k = 5.3 N/m. At t = 0 s, the block is 20.0 cm below the equilibrium position and moving upward with a speed of 194 cm/s. What is the block's speed when the displacement from equilibrium is 31.5 cm?
ω=sqrt(k/m)=sqrt(4.6/0.1) = 6.76rad/s.
x=A•sinωt
v=dx/dt=A•ω•cosωt
Divide the first equation by the second and obtain
x/v= A•sinωt/ A•ω•cosωt= ω•tanωt,
tanωt = x•ω/v=20•6.76/194=0.7
ωt =arctanωt =0.61 rad,
sinωt = 0.571
A=x/sinωt = 20/0.571=35 cm.
For the second case:
x1=A•sinωt1,
sinωt1 =x1/A=31.5/35 =0.9.
cos ωt1=sqrt(1-sin²ωt1)=0.44.
v1= =A•ω•cosωt1=
=35•6.76•0.44=104 cm/s.
K is 5.3 N/m. Why are you using 4.6 N/m?
This is the problem similar to that I've solved for another post. Substitute your data and calculate yourself
Check my calculations.
ω=sqrt(k/m)=sqrt(5.3/0.1) = 7.28rad/s.
x=A•sinωt
v=dx/dt=A•ω•cosωt
Divide the first equation by the second and obtain
x/v= A•sinωt/ A•ω•cosωt= ω•tanωt,
tanωt = x•ω/v=20•7.28/194=0.75
ωt =arctanωt =0.64 rad,
sinωt = 0.6
A=x/sinωt = 20/0.571=33.3 cm.
For the second case:
x1=A•sinωt1,
sinωt1 =x1/A=31.5/33.3 =0.95.
cos ωt1=sqrt(1-sin²ωt1)=0.33.
v1= =A•ω•cosωt1=
=33.3•7.28•0.33=80 cm/s.
I got a similar answer of 79.0 cm/s using the conservation of energy equation.
U(i) + K(i) = U(f) + U(f)
Thank you for the help. And sorry for asking this question. I didn't see the similar question when I searched mine in the search bar.
To find the block's speed when the displacement from equilibrium is 31.5 cm, we can first calculate the potential energy and kinetic energy at the initial position and then use the conservation of mechanical energy.
1. Find the potential energy at the initial position:
The potential energy of a spring is given by the formula U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the displacement from the equilibrium position is 20.0 cm below, so x = -20.0 cm.
U_initial = (1/2) × k × x^2
= (1/2) × 5.3 N/m × (-20.0 cm)^2
= 10.6 N/m × 400 cm^2
= 4240 J
2. Find the kinetic energy at the initial position:
The kinetic energy is given by the formula K = (1/2)mv^2, where K is the kinetic energy, m is the mass, and v is the velocity.
In this case, the mass (m) is given as 100 g, which is 0.1 kg, and the velocity (v) is given as 194 cm/s.
K_initial = (1/2) × 0.1 kg × (194 cm/s)^2
= 0.05 kg × 37636 cm^2/s^2
= 1881.8 J
3. Calculate the total mechanical energy at the initial position:
The total mechanical energy is the sum of potential energy and kinetic energy.
E_initial = U_initial + K_initial
= 4240 J + 1881.8 J
= 6121.8 J
4. Calculate the velocity at the final position:
To find the velocity at the final position, we use the conservation of mechanical energy: E_initial = E_final.
At the final position, the displacement from the equilibrium is 31.5 cm above, so x = 31.5 cm.
E_final = U_final + K_final
Since the block is at the highest point (maximum potential energy), the kinetic energy at the final position is zero, so K_final = 0.
E_final = U_final + 0
U_final = E_initial - U_final
From the equation for potential energy, we can rearrange it to solve for the displacement:
U = (1/2)kx^2
(x^2) = (2U) / k
x = sqrt((2U) / k)
Substituting the values:
x = sqrt((2 × 6121.8 J) / 5.3 N/m)
x = sqrt(2307.2 cm^2)
x = 48 cm
Therefore, at a displacement of 31.5 cm, the block's displacement (x) is 48 cm above the equilibrium position.
5. Find the final potential energy:
U_final = (1/2) × k × x^2
= (1/2) × 5.3 N/m × (48 cm)^2
= 12.7 × 2304 J
= 29,068.8 J
6. Calculate the final kinetic energy:
Since the block is at the highest point and momentarily stops before starting to descend, the kinetic energy is zero, so K_final = 0.
7. Find the final velocity using the conservation of mechanical energy:
E_final = U_final + K_final
= 29,068.8 J + 0
= 29,068.8 J
From the formula for kinetic energy, we can rearrange it to solve for velocity:
K = (1/2)mv^2
v^2 = (2K) / m
v = sqrt((2K) / m)
Substituting the values:
v = sqrt((2 × 29,068.8 J) / 0.1 kg)
v = sqrt(581,376 J/kg)
v ≈ 241.1 m/s
Therefore, the block's speed when the displacement from equilibrium is 31.5 cm is approximately 241.1 m/s.