Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 186 N/m. The motion of the object on spring 1 has 3 times the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.
My work...
Aw=Aw
A(sqrt of k/m) = A(sqrt k/m)
(sqrt of 186) = 3 (sqrt of k)
(sqrt of 186)/3 = (sqrt of k)
[(sqrt of 186)/3] ^2 = k
I got 20.6 but this is wrong, Please help!
body 1
x = sin 2 pi f t
v = 2 pi f cos 2 pi f t
a = -(2pi)^2 f^2 sin 2 pi f t
-k x = m a
- k x = - m(2pi)^2 f^2 sin 2 pi f t
- k/m = - (2 pi)^2 f^2
f = (1/2pi) sqrt(k/m)
body 2
x = 3 sin 2 pi F t
v = 3 (2 pi) F cos 2 pi F t
- K/m = -3 (2 pi)^2 F^2 sin 2 pi F t
-K/m =-(2 pi)^2 F^2
F = (1/2pi) sqrt(K/m)
the to v values are equal in magnitude
2 pi f = 6 pi F
2 pi ((1/2pi) sqrt(k/m)= 6 pi(1/2pi) sqrt(K/m)
sqrt(k/m) = 3 sqrt(K/m)
k/m = 9 K/m
K = k/9 = 20.67
To solve this problem, let's follow these steps:
1. Express the given information in mathematical form:
Let:
- k1 be the spring constant of spring 1.
- k2 be the spring constant of spring 2.
- A1 be the amplitude of the motion of the object on spring 1.
- A2 be the amplitude of the motion of the object on spring 2.
Given:
A1 = 3A2 (the amplitude of motion on spring 1 is three times the amplitude of motion on spring 2)
Also, the magnitude of the maximum velocity (v) is the same in both cases.
2. Use the formula for velocity in simple harmonic motion:
v = ωA
where v is the magnitude of the maximum velocity, ω is the angular frequency, and A is the amplitude.
Since the magnitude of the maximum velocity is the same in both cases, we can write:
ω1A1 = ω2A2
3. Substitute the expressions for A1 and A2 in terms of ω1 and ω2:
ω1(3A2) = ω2A2
Simplify:
3ω1 = ω2
4. Use the formula for angular frequency in terms of the spring constant and mass:
ω = √(k/m)
where ω is the angular frequency, k is the spring constant, and m is the mass.
5. Square both sides of the equation ω1 = ω2/3 to get rid of the square root:
(ω1)^2 = (ω2/3)^2
6. Substitute the expressions for angular frequency in terms of the spring constant:
(k1/m)(A1^2) = (k2/m)(A2^2)/9
7. Cancel out the mass terms:
k1A1^2 = k2A2^2/9
8. Substitute the given values:
186(A1^2) = k2(A2^2)/9
9. Use the relationship A1 = 3A2:
186(9A2^2) = k2(A2^2)/9
10. Simplify the equation:
1674A2^2 = k2A2^2
11. Divide both sides by A2^2 to isolate k2:
1674 = k2
Therefore, the spring constant of spring 2 is 1674 N/m.