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A mass m is suspended from a spring of stiffness 4000 N/m and is subjected to a harmonic force having an amplitude of 100N and a frequency of 5Hz. The amplitude of the forced motion of mass is observed to be 20mm. The value of m is nearly
- a)9 Kg
- b)360 Kg
- c)20 Kg
- d)190 Kg
Correct answer is option 'A'. Can you explain this answer?
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A mass m is suspended from aspring of stiffness 4000 N/m andis subject...
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A mass m is suspended from aspring of stiffness 4000 N/m andis subject...
Given:
Stiffness of the spring, k = 4000 N/m
Amplitude of harmonic force, F0 = 100 N
Frequency of harmonic force, ω = 2πf = 2π × 5 = 31.42 rad/s
Amplitude of forced motion, x0 = 20 mm = 0.02 m
Formula used:
The equation of motion of a mass-spring system under a harmonic force is given by:
mx″ + kx = F0cos(ωt)
where,
m = mass
x = displacement
x″ = acceleration
Solution:
1. We can find the mass of the object using the equation of motion. Substituting the given values in the equation, we get:
mx″ + kx = F0cos(ωt)
m(ω^2)x + kx = F0cos(ωt)
At maximum displacement, x = x0 = 0.02 m and cos(ωt) = 1.
Therefore, we get:
m(ω^2)x0 + kx0 = F0
m = (F0/k) - (ω^2)x0
m = (100/4000) - (31.42^2) × 0.02
m = 9.02 kg ≈ 9 kg
2. Hence, the value of m is nearly 9 kg.
Therefore, option (a) is the correct answer.
Stiffness of the spring, k = 4000 N/m
Amplitude of harmonic force, F0 = 100 N
Frequency of harmonic force, ω = 2πf = 2π × 5 = 31.42 rad/s
Amplitude of forced motion, x0 = 20 mm = 0.02 m
Formula used:
The equation of motion of a mass-spring system under a harmonic force is given by:
mx″ + kx = F0cos(ωt)
where,
m = mass
x = displacement
x″ = acceleration
Solution:
1. We can find the mass of the object using the equation of motion. Substituting the given values in the equation, we get:
mx″ + kx = F0cos(ωt)
m(ω^2)x + kx = F0cos(ωt)
At maximum displacement, x = x0 = 0.02 m and cos(ωt) = 1.
Therefore, we get:
m(ω^2)x0 + kx0 = F0
m = (F0/k) - (ω^2)x0
m = (100/4000) - (31.42^2) × 0.02
m = 9.02 kg ≈ 9 kg
2. Hence, the value of m is nearly 9 kg.
Therefore, option (a) is the correct answer.
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A mass m is suspended from aspring of stiffness 4000 N/m andis subjected to a harmonic forcehaving an amplitude of 100N and afrequency of 5Hz. The amplitude ofthe forced motion of mass isobserved to be 20mm. The value ofm is nearlya)9 Kgb)360 Kgc)20 Kgd)190 KgCorrect answer is option 'A'. Can you explain this answer?
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