I keep getting this question wrong. I would really appreciate some guidance.
The maximum speed of a 3.6 kg mass attached to a spring is 0.69 m/s, and the maximum force exerted on the mass is 11 N.
(1) What is the amplitude of motion for this mass?
(2) What is the force constant of the spring?
(3) What is the frequency of this system?
The maximum speed of a 3.6 kg mass attached to a spring is 0.69 m/s, and the maximum force exerted on the mass is 11 N.
(1) What is the amplitude of motion for this mass?
(2) What is the force constant of the spring?
(3) What is the frequency of this system?
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It is supposed that the mass attached to the spring is vibrating with SHM
Mass of the vibrating body = m, acceleration at any instant = a
In an SHM, Force acting on the vibrating mass = F, Displacement at any instant = x
F = - kx , where k is a contant
m a = - k x , => a = - (k/m) x = - ω² x, where ω² = k/m and therefore k = mω²
m = 3.6 kg
Maximum displacement = Amplitude = A
Maximum force (magnitude only) = F(max) = k A = mω²A = 11 N ........(1)
Maximum acceleration (magnitude only) = a(max) = (k/m) A = ω²A ...... (2)
Maximum speed = V(max) = ω A = 0.69 m/s .......(3)
From (1) and (3) F(max) / V(max) = m ω = 11 / (0.69)
=> 3.6 ω = 11 / (0.69), => ω = 11 / {(0.69)*(3.6)} = 4.43 ..... (4)
From (3) and (4) , A = 0.69 / 4.43 = 0.156 m ...... (5)
Force constant = k = F(max) / A = 11 / (0.156) = 70.51 N/m ..... (6)
ω = 2 π f, where f = frequency of the system
=> f = ω / 2π = 4.43 / 6.28 = 0.71 Hz .....(7)
Answers are found in (5), (6) and (7)
Mass of the vibrating body = m, acceleration at any instant = a
In an SHM, Force acting on the vibrating mass = F, Displacement at any instant = x
F = - kx , where k is a contant
m a = - k x , => a = - (k/m) x = - ω² x, where ω² = k/m and therefore k = mω²
m = 3.6 kg
Maximum displacement = Amplitude = A
Maximum force (magnitude only) = F(max) = k A = mω²A = 11 N ........(1)
Maximum acceleration (magnitude only) = a(max) = (k/m) A = ω²A ...... (2)
Maximum speed = V(max) = ω A = 0.69 m/s .......(3)
From (1) and (3) F(max) / V(max) = m ω = 11 / (0.69)
=> 3.6 ω = 11 / (0.69), => ω = 11 / {(0.69)*(3.6)} = 4.43 ..... (4)
From (3) and (4) , A = 0.69 / 4.43 = 0.156 m ...... (5)
Force constant = k = F(max) / A = 11 / (0.156) = 70.51 N/m ..... (6)
ω = 2 π f, where f = frequency of the system
=> f = ω / 2π = 4.43 / 6.28 = 0.71 Hz .....(7)
Answers are found in (5), (6) and (7)