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A single degree of freedom system having mass 1 kg and stiffness 10kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10−4 seconds. The amplitude in mm of the resulting free vibration is
- a)0.5
- b)1.0
- c)5.0
- d)10.0
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A single degree of freedom system having mass 1 kg and stiffness 10kN/...
Milliseconds. Find the displacement and velocity of the system after the impulse.
We can use the impulse-momentum equation to solve this problem:
impulse = change in momentum
The impulse is given by the magnitude of the force multiplied by the duration of the force:
impulse = 5 kN * 10 ms = 50 Ns
The initial momentum of the system is zero, since it is initially at rest. Therefore, the change in momentum is equal to the final momentum:
change in momentum = final momentum
The final momentum can be expressed in terms of the displacement and velocity of the system:
final momentum = m * v = 1 kg * v
where v is the velocity of the system after the impulse.
The displacement can be found using the equation of motion for a single degree of freedom system:
m * x'' + k * x = f(t)
where x'' is the second derivative of displacement with respect to time, k is the stiffness, and f(t) is the applied force. Since the system is initially at rest, we can assume that the displacement at t=0 is zero. The impulse force can be modeled as a delta function, so f(t) = 5 kN * delta(t-10 ms).
Using Laplace transforms, we can solve for the displacement:
s^2 * X(s) + 10^4 * X(s) = 5 * 10^3 * e^(-10s)
X(s) = 5 * 10^3 / (s^2 + 10^4) * e^(-10s)
Taking the inverse Laplace transform, we get:
x(t) = 5 * 10^3 / 10^4 * (1 - e^(-10t) * cos(10^4 t))
At t=10 ms, the displacement is:
x(10 ms) = 5 * 10^3 / 10^4 * (1 - e^(-1) * cos(100)) ≈ 0.305 mm
To find the velocity, we can take the derivative of the displacement:
v(t) = dx/dt = 5 * 10^3 / 10^4 * (10e^(-10t) * sin(10^4 t) - e^(-10t) * 10^4 * cos(10^4 t))
At t=10 ms, the velocity is:
v(10 ms) = 5 * 10^3 / 10^4 * (10e^(-1) * sin(100) - e^(-1) * 10^4 * cos(100)) ≈ -4.269 m/s
Therefore, the displacement is approximately 0.305 mm and the velocity is approximately -4.269 m/s after the impulse.
We can use the impulse-momentum equation to solve this problem:
impulse = change in momentum
The impulse is given by the magnitude of the force multiplied by the duration of the force:
impulse = 5 kN * 10 ms = 50 Ns
The initial momentum of the system is zero, since it is initially at rest. Therefore, the change in momentum is equal to the final momentum:
change in momentum = final momentum
The final momentum can be expressed in terms of the displacement and velocity of the system:
final momentum = m * v = 1 kg * v
where v is the velocity of the system after the impulse.
The displacement can be found using the equation of motion for a single degree of freedom system:
m * x'' + k * x = f(t)
where x'' is the second derivative of displacement with respect to time, k is the stiffness, and f(t) is the applied force. Since the system is initially at rest, we can assume that the displacement at t=0 is zero. The impulse force can be modeled as a delta function, so f(t) = 5 kN * delta(t-10 ms).
Using Laplace transforms, we can solve for the displacement:
s^2 * X(s) + 10^4 * X(s) = 5 * 10^3 * e^(-10s)
X(s) = 5 * 10^3 / (s^2 + 10^4) * e^(-10s)
Taking the inverse Laplace transform, we get:
x(t) = 5 * 10^3 / 10^4 * (1 - e^(-10t) * cos(10^4 t))
At t=10 ms, the displacement is:
x(10 ms) = 5 * 10^3 / 10^4 * (1 - e^(-1) * cos(100)) ≈ 0.305 mm
To find the velocity, we can take the derivative of the displacement:
v(t) = dx/dt = 5 * 10^3 / 10^4 * (10e^(-10t) * sin(10^4 t) - e^(-10t) * 10^4 * cos(10^4 t))
At t=10 ms, the velocity is:
v(10 ms) = 5 * 10^3 / 10^4 * (10e^(-1) * sin(100) - e^(-1) * 10^4 * cos(100)) ≈ -4.269 m/s
Therefore, the displacement is approximately 0.305 mm and the velocity is approximately -4.269 m/s after the impulse.
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Community Answer
A single degree of freedom system having mass 1 kg and stiffness 10kN/...
Wn=√10000/1
=100rad/sec
initial velocity is 0
F*t=m(Vo-Vi)
(5*10^3*10^-4)=1(Vo-0)
0.5=1(Vo)
Vo=0.5m/s
X=Vo/Wn
=0.5/100
=0.005 m OR
=5 mm
=100rad/sec
initial velocity is 0
F*t=m(Vo-Vi)
(5*10^3*10^-4)=1(Vo-0)
0.5=1(Vo)
Vo=0.5m/s
X=Vo/Wn
=0.5/100
=0.005 m OR
=5 mm
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A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer?
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A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer?.
A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single degree of freedom system having mass 1 kg and stiffness 10kN/minitially at rest is subjected to an impulse force of magnitude 5 kN for10−4 seconds. The amplitude in mm of the resulting free vibration isa)0.5b)1.0c)5.0d)10.0Correct answer is option 'C'. Can you explain this answer?.
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