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A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?
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?A point mass m=20kg, is suspended by a massless spring of constant 20...
The equation of displacement of a point mass suspended by a spring can be derived using the principles of simple harmonic motion. Let's analyze the given scenario step by step:

1. Initial conditions:
- Mass of the point mass, m = 20 kg
- Spring constant, k = 2000 N/m
- Elongation in the spring, x = 15 cm = 0.15 m
- Acceleration due to gravity, g = 10 m/s²

2. Equilibrium position:
- When the point mass is at rest, it is in equilibrium. At this position, the net force acting on the mass is zero.
- The force exerted by the spring, Fs = -kx, where x is the displacement from the equilibrium position.
- The weight of the mass, Fg = mg, acts downwards.
- At equilibrium, Fs = Fg, so -kx = mg.

3. Equation of motion:
- The equation of motion for a mass-spring system is given by m(d²y/dt²) + k(y - y_eq) = 0, where y represents displacement and y_eq is the equilibrium position.
- Rearranging the equation, we get d²y/dt² + (k/m)y = (k/m)y_eq.

4. Solving the equation:
- Substituting the values, we have d²y/dt² + (2000/20)y = (2000/20)(-0.15).
- Simplifying the equation, we get d²y/dt² + 100y = -15.

5. Solving the differential equation:
- The characteristic equation for this differential equation is r² + 100 = 0.
- Solving the characteristic equation, we find two imaginary roots: r = ±10i.

6. General solution:
- The general solution for this differential equation is y(t) = A*cos(10t) + B*sin(10t), where A and B are constants.
- Since the mass is released from the elongated position, the initial velocity is zero. Therefore, we can ignore the cosine term (which represents the initial velocity) in the general solution.
- Thus, the equation of displacement becomes y(t) = B*sin(10t).

7. Determining the constant B:
- To determine the value of B, we need to consider the initial displacement. Given that the elongation in the spring is 15 cm, we have y(t = 0) = 0.15 m.
- Substituting the values, we get 0.15 = B*sin(10*0).
- Since sin(0) = 0, B must be equal to 0.15.

8. Final equation of displacement:
- Substituting the value of B, the equation of displacement becomes y(t) = 0.15*sin(10t).

Therefore, the correct equation of displacement as a function of time is y = 0.15*sin(10t), which is option D.
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?A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?
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?A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about ?A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?.
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