JEE Main & Advanced Numerical Shortcut methods and Tricks

1. A block of mass 2 kg is attached to a spring with a spring constant of 100 N/m. The block is pulled 10 cm to the right of its equilibrium position and released. Determine the maximum velocity of the block.

Solution:

• The restoring force can be calculated as $$F = -kx = -100 \times 0.1 = -10 N$$
• The maximum velocity can be determined using the formula: $$v_{max} = \sqrt{k/m} \times x = \sqrt{100/2} \times 0.1 = 1 m/s$$

2. A 20 kg block is suspended from a spring with a spring constant of 500 N/m. The block is pulled down 10 cm from its equilibrium position and released. Determine the period of oscillation.

Solution:

• The period of oscillation can be determined using the formula: $$T = 2\pi\sqrt{m/k} = 2\pi \sqrt{20/500} = 0.63 s$$

3. Two blocks of masses 1 kg each are connected by a spring of spring constant 100 N/m. The blocks are placed on a horizontal frictionless surface and the spring is stretched by 10 cm. Determine the speed of each block when they are released.

Solution:

• The initial elastic potential energy stored in the spring is $$U = 1/2kx^2 = 1/2 \times 100 \times (0.1)^2 = 0.5 J$$
• When the blocks are released, the potential energy is converted into kinetic energy for each block $$K = 1/2mv2 = 0.5 J => v = 0.5 m/s$$

CBSE Board Exam Numerical

1. A block of mass 1 kg is attached to a vertical spring. The spring constant is 100 N/m. The block is pulled down 10 cm below its equilibrium position and released. Determine the maximum height reached by the block.

Solution:

• The maximum height reached by the block can be determined using conservation of energy: $$h = 1/2 (v^2/g) = (1/2)(2gh) = 0.5 m$$

2. A spring with a spring constant of 200 N/m is stretched by 10 cm from its equilibrium position. A ball of mass 0.5 kg is released from rest at this point. Determine the velocity of the ball when it passes through the equilibrium position.

Solution:

• Using conservation of energy, $$1/2kx^2 = 1/2mv^2 => v = \sqrt{(kx^2)/m} = 2 m/s$$

3. A block of mass 1 kg is attached to a spring with a spring constant of 100 N/m. The block is pulled 10 cm from its equilibrium position and held at rest. Determine the energy stored in the spring.

Solution:

• The energy stored in the spring can be calculated using the formula $$U = 1/2kx^2 = 1/2 \times 100 \times (0.1)^2 = 0.5 J$$