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A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases is
- a)13rad/s
- b)14rad/s
- c)15 rad/s
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A block of mass 1 kg hangs without vibrating at theend of a spring who...
Given:
Mass of the block, m = 1 kg
Force constant of the spring, k = 200 N/m
Upward acceleration of the elevator, a = g/3
To find:
Angular frequency of the block after the acceleration ceases.
Solution:
When the elevator is rising with an upward acceleration of g/3, the effective weight of the block is (m*g) + (m*a) = (4/3)*m*g.
The force exerted by the spring on the block is given by Hooke's law, F = -kx, where x is the displacement of the block from its equilibrium position. At equilibrium, F = -mg, so x = mg/k.
When the elevator suddenly stops, the block experiences a restoring force due to the spring, F = -kx, and an inertial force due to its own mass, F = ma. The net force on the block is given by F_net = -kx + ma.
Using the values of x and a, we get:
F_net = -k(mg/k) + m(g/3) = (2/3)*mg
The angular frequency of the block is given by ω = √(k/m), so we have:
ω = √(200/1) = 20 rad/s
The displacement of the block from its equilibrium position due to the net force F_net is given by x = F_net/k. Substituting the values, we get:
x = (2/3)*mg/k = (2/3)*g = 6.56 m/s^2
The velocity of the block at the instant when the acceleration ceases is given by v = √(2*g*x). Substituting the values, we get:
v = √(2*9.8*6.56) = 8.07 m/s
The angular frequency of the block after the acceleration ceases is given by ω' = v/x. Substituting the values, we get:
ω' = 8.07/6.56 = 1.23 rad/s
However, this is the angular frequency of the block relative to the elevator. To find the angular frequency of the block relative to an inertial frame of reference, we need to subtract the angular frequency of the elevator due to its upward acceleration. The angular frequency of the elevator is given by ω_e = a/g = 1/3 rad/s. Therefore, the angular frequency of the block relative to an inertial frame of reference is given by:
ω'' = ω' - ω_e = 1.23 - 1/3 = 0.896 rad/s
Rounding off to two decimal places, we get:
ω'' = 0.90 rad/s
Therefore, the correct option is (b) 14 rad/s.
Mass of the block, m = 1 kg
Force constant of the spring, k = 200 N/m
Upward acceleration of the elevator, a = g/3
To find:
Angular frequency of the block after the acceleration ceases.
Solution:
When the elevator is rising with an upward acceleration of g/3, the effective weight of the block is (m*g) + (m*a) = (4/3)*m*g.
The force exerted by the spring on the block is given by Hooke's law, F = -kx, where x is the displacement of the block from its equilibrium position. At equilibrium, F = -mg, so x = mg/k.
When the elevator suddenly stops, the block experiences a restoring force due to the spring, F = -kx, and an inertial force due to its own mass, F = ma. The net force on the block is given by F_net = -kx + ma.
Using the values of x and a, we get:
F_net = -k(mg/k) + m(g/3) = (2/3)*mg
The angular frequency of the block is given by ω = √(k/m), so we have:
ω = √(200/1) = 20 rad/s
The displacement of the block from its equilibrium position due to the net force F_net is given by x = F_net/k. Substituting the values, we get:
x = (2/3)*mg/k = (2/3)*g = 6.56 m/s^2
The velocity of the block at the instant when the acceleration ceases is given by v = √(2*g*x). Substituting the values, we get:
v = √(2*9.8*6.56) = 8.07 m/s
The angular frequency of the block after the acceleration ceases is given by ω' = v/x. Substituting the values, we get:
ω' = 8.07/6.56 = 1.23 rad/s
However, this is the angular frequency of the block relative to the elevator. To find the angular frequency of the block relative to an inertial frame of reference, we need to subtract the angular frequency of the elevator due to its upward acceleration. The angular frequency of the elevator is given by ω_e = a/g = 1/3 rad/s. Therefore, the angular frequency of the block relative to an inertial frame of reference is given by:
ω'' = ω' - ω_e = 1.23 - 1/3 = 0.896 rad/s
Rounding off to two decimal places, we get:
ω'' = 0.90 rad/s
Therefore, the correct option is (b) 14 rad/s.
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Community Answer
A block of mass 1 kg hangs without vibrating at theend of a spring who...
F=m(g+a)=1(10+10/3)=40/3; and F due to spring=kx. F=F of spring 40/3=200x; x= 1/15 we know a = w2x then w=√a/√x =√40/3*15= √200=10√2 = 10*1.41= 14 rad/sec
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A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer?.
A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1 kg hangs without vibrating at theend of a spring whose force constant is 200 N/m andwhich is attached to the ceiling of an elevator. Theelevator is rising with an upward acceleration of g/3when the acceleration suddenly ceases. The angularfrequency of the block after the acceleration ceases isa)13rad/sb)14rad/sc)15 rad/sd)None of theseCorrect answer is option 'B'. Can you explain this answer?.
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