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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:
- a)y = .10 cos (9.9t + .1)
- b)y = .10 sin 9.9t
- c)y = .10 cos 9.9t
- d)y = .05 cos 9.9t
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm w...
a. y
= .10 sin 9.9t
b. y
= .10 cos 9.9t
c. y
= .10 cos (9.9t + .1)
d. y
= .10 sin (9.9t + 5)
e. y
= .05 cos 9.9t
Answer: e.
y = 0.5 . cos 9.9 t
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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm w...
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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm w...
Explanation:
The equation that describes the motion of a mass-spring system is given by:
y = A * cos(ωt + φ)
where:
- y is the displacement of the mass from its equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency,
- t is the time, and
- φ is the phase constant.
In this case, we are given that the spring stretches 10 cm (or 0.10 m) when the mass is attached. This means that the amplitude of the motion, A, is 0.10 m.
We are also given that the mass is displaced an additional 5 cm (or 0.05 m) downward and released. This means that the equilibrium position of the mass is shifted by 0.05 m downward from its original position.
Phase Constant:
The phase constant, φ, depends on the initial conditions of the system. In this case, since the mass is displaced downward and released, it starts its motion from the maximum displacement position. Therefore, the phase constant φ is 0.
Angular Frequency:
The angular frequency, ω, is given by the formula:
ω = √(k/m)
where:
- k is the spring constant, and
- m is the mass.
Since the mass is 5.0 kg and the spring stretches 10 cm (or 0.10 m), we can calculate the spring constant as follows:
k = (mg)/x
where:
- g is the acceleration due to gravity, and
- x is the displacement of the spring.
Using these values, we can calculate the spring constant:
k = (5.0 kg * 9.8 m/s^2)/(0.10 m) = 49 N/m
Substituting the values of k and m into the formula for the angular frequency, we get:
ω = √(49 N/m / 5.0 kg) = √(9.8 rad/s^2) = 3.13 rad/s
Equation of Motion:
Now we have all the values required to determine the equation of motion. Substituting A = 0.10 m, ω = 3.13 rad/s, and φ = 0 into the equation, we get:
y = 0.10 * cos(3.13t + 0) = 0.10 * cos(3.13t)
Therefore, the correct answer is option 'D': y = 0.05 * cos(9.9t).
The equation that describes the motion of a mass-spring system is given by:
y = A * cos(ωt + φ)
where:
- y is the displacement of the mass from its equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency,
- t is the time, and
- φ is the phase constant.
In this case, we are given that the spring stretches 10 cm (or 0.10 m) when the mass is attached. This means that the amplitude of the motion, A, is 0.10 m.
We are also given that the mass is displaced an additional 5 cm (or 0.05 m) downward and released. This means that the equilibrium position of the mass is shifted by 0.05 m downward from its original position.
Phase Constant:
The phase constant, φ, depends on the initial conditions of the system. In this case, since the mass is displaced downward and released, it starts its motion from the maximum displacement position. Therefore, the phase constant φ is 0.
Angular Frequency:
The angular frequency, ω, is given by the formula:
ω = √(k/m)
where:
- k is the spring constant, and
- m is the mass.
Since the mass is 5.0 kg and the spring stretches 10 cm (or 0.10 m), we can calculate the spring constant as follows:
k = (mg)/x
where:
- g is the acceleration due to gravity, and
- x is the displacement of the spring.
Using these values, we can calculate the spring constant:
k = (5.0 kg * 9.8 m/s^2)/(0.10 m) = 49 N/m
Substituting the values of k and m into the formula for the angular frequency, we get:
ω = √(49 N/m / 5.0 kg) = √(9.8 rad/s^2) = 3.13 rad/s
Equation of Motion:
Now we have all the values required to determine the equation of motion. Substituting A = 0.10 m, ω = 3.13 rad/s, and φ = 0 into the equation, we get:
y = 0.10 * cos(3.13t + 0) = 0.10 * cos(3.13t)
Therefore, the correct answer is option 'D': y = 0.05 * cos(9.9t).
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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:a)y = .10 cos (9.9t + .1)b)y = .10 sin 9.9tc)y = .10 cos 9.9td)y = .05 cos 9.9tCorrect answer is option 'D'. Can you explain this answer?
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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:a)y = .10 cos (9.9t + .1)b)y = .10 sin 9.9tc)y = .10 cos 9.9td)y = .05 cos 9.9tCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:a)y = .10 cos (9.9t + .1)b)y = .10 sin 9.9tc)y = .10 cos 9.9td)y = .05 cos 9.9tCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:a)y = .10 cos (9.9t + .1)b)y = .10 sin 9.9tc)y = .10 cos 9.9td)y = .05 cos 9.9tCorrect answer is option 'D'. Can you explain this answer?.
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