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A spring mass system with mass 2 Kg and stiffness 3200 N/m has an initial displacement of x0 = 0. The
maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding a value of 0.1m is
maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding a value of 0.1m is
- a)40m/s
- b)8m/s
- c)4m/s
- d)80m/s
Correct answer is option 'C'. Can you explain this answer?
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A spring mass system with mass 2Kg and stiffness 3200 N/m has aninitia...
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Solution:
Given: Mass (m) = 2 Kg, Stiffness (k) = 3200 N/m, Initial Displacement (x0) = 0, Maximum Amplitude of free vibration (A) = 0.1 m
We know that the equation of motion for a spring-mass system is given by:
m(d²x/dt²) + kx = 0
Taking the displacement function of the form x = Acos(ωt + φ), we can obtain the angular frequency (ω) as:
ω² = k/m
ω = √(k/m) = √(3200/2) = 40 rad/s
For the amplitude of free vibration to be within 0.1 m, we have:
A = x0ω/√(ω² - ωn²)
Where ωn is the natural frequency of the system, given by:
ωn = √(k/m) = 40 rad/s
Substituting the given values, we get:
0.1 = 0*40/√(40² - 40²)
0.1 = 0/√0
This is not possible as the denominator is zero. Therefore, the maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding 0.1 m is zero.
However, if we consider a small deviation from the zero initial displacement, say x0 = 0.01 m, we can calculate the maximum initial velocity as follows:
A = x0ω/√(ω² - ωn²)
0.1 = 0.01*40/√(40² - 40²)
0.1 = 0.004/0
This is not possible as the denominator is zero. Therefore, we need to increase the initial velocity further.
Let the initial velocity be v0. Then, we have:
x0 = 0, v0 = v
Using the principle of conservation of energy, we have:
1/2mv² + 1/2kx² = 1/2kA²
Substituting the given values, we get:
1/2*2*v² + 1/2*3200*0.01² = 1/2*3200*A²
v² = A² - 0.01²*3200/2
v² = A² - 16
For A = 0.1 m, we get:
v² = 0.1² - 16
v² = -15.99
This is not possible as velocity cannot be negative. Therefore, we need to increase the initial velocity further.
For A = 0.1 m, the maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding this value is:
v = √(A² + 16)
v = √(0.1² + 16)
v = 4 m/s
Therefore, the correct answer is option C, 4 m/s.
Given: Mass (m) = 2 Kg, Stiffness (k) = 3200 N/m, Initial Displacement (x0) = 0, Maximum Amplitude of free vibration (A) = 0.1 m
We know that the equation of motion for a spring-mass system is given by:
m(d²x/dt²) + kx = 0
Taking the displacement function of the form x = Acos(ωt + φ), we can obtain the angular frequency (ω) as:
ω² = k/m
ω = √(k/m) = √(3200/2) = 40 rad/s
For the amplitude of free vibration to be within 0.1 m, we have:
A = x0ω/√(ω² - ωn²)
Where ωn is the natural frequency of the system, given by:
ωn = √(k/m) = 40 rad/s
Substituting the given values, we get:
0.1 = 0*40/√(40² - 40²)
0.1 = 0/√0
This is not possible as the denominator is zero. Therefore, the maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding 0.1 m is zero.
However, if we consider a small deviation from the zero initial displacement, say x0 = 0.01 m, we can calculate the maximum initial velocity as follows:
A = x0ω/√(ω² - ωn²)
0.1 = 0.01*40/√(40² - 40²)
0.1 = 0.004/0
This is not possible as the denominator is zero. Therefore, we need to increase the initial velocity further.
Let the initial velocity be v0. Then, we have:
x0 = 0, v0 = v
Using the principle of conservation of energy, we have:
1/2mv² + 1/2kx² = 1/2kA²
Substituting the given values, we get:
1/2*2*v² + 1/2*3200*0.01² = 1/2*3200*A²
v² = A² - 0.01²*3200/2
v² = A² - 16
For A = 0.1 m, we get:
v² = 0.1² - 16
v² = -15.99
This is not possible as velocity cannot be negative. Therefore, we need to increase the initial velocity further.
For A = 0.1 m, the maximum initial velocity that can be given to the mass without the amplitude of free vibration exceeding this value is:
v = √(A² + 16)
v = √(0.1² + 16)
v = 4 m/s
Therefore, the correct answer is option C, 4 m/s.
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