Typical Numericals on Problem-Solving: Simple Harmonic Motion JEE Main Exam:

1. A mass-spring system:

• a) Amplitude: $$0.1 \text{ m}$$
• b) Angular frequency: $$10 \text{ rad/s}$$
• c) Maximum velocity: $$0.1 \text{ m/s}$$

2. A simple pendulum:

• a) Time period: $$2 \pi \sqrt{\frac{1}{9.81}} \approx 2.01 \text{ s}$$
• b) Maximum angular velocity: $$\omega_{max}=\sqrt{\frac{g}{l}} \sin\theta_0=\sqrt{\frac{9.81}{1}} \sin10^\circ \approx 0.174 \text{ rad/s}$$
• c) Maximum potential energy: $$U_{max}=mgl\left(1-\cos\theta_0\right)=1 \times 9.81 \times 1 (1-\cos10^\circ) \approx 0.173 \text{ J}$$

3. Particle in SHM:

• a) Maximum velocity: $$2\pi f A= 2\pi \times 2 \times 0.2 = 2.51 \text{ m/s}$$
• b) Maximum acceleration: $$\omega^2 A= (2\pi f)^2 A = (2\pi \times 2)^2 \times 0.2 = 31.42 \text{ m/s}^2$$
• c) Total energy: $$E=\frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2=3.14 \text{ J}$$

4. Block-spring system:

• a) Time period: $$T=2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2}{10}} \approx 2.83 \text{ s}$$
• b) Angular frequency: $$\omega= \sqrt{\frac{k}{m}}=\sqrt{\frac{10}{2}}= 2.24 \text{ rad/s}$$
• c) Maximum velocity: $$\omega A = 2.24 \times 0.5=1.12 \text{ m/s}$$

CBSE Board Exam:

1. Mass-spring system:

• a) Amplitude: $$0.2 \text{ m}$$
• b) Time period: $$2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{1}{100}} \approx 0.63 \text{ s}$$
• c) Maximum velocity: $$A\omega= 0.2 \times \sqrt{\frac{100}{1}} = 2 \text{ m/s}$$

2. Simple pendulum:

• a) Time period: $$2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{2}{9.81}} \approx 2.83 \text{ s}$$
• b) Maximum angular velocity: $$\omega_{max}=\sqrt{\frac{g}{l}} \sin\theta_0=\sqrt{\frac{9.81}{2}} \sin15^\circ \approx 0.54 \text{ rad/s}$$
• c) Maximum potential energy: $$U_{max}=mgl\left(1-\cos\theta_0\right)=1 \times 9.81 \times 2 (1-\cos15^\circ) \approx 1.07 \text{ J}$$

3. Particle in SHM:

• a) Maximum velocity: $$2\pi f A= 2\pi \times 5 \times 0.3 = 9.42 \text{ m/s}$$
• b) Maximum acceleration: $$\omega^2 A= (2\pi f)^2 A = (2\pi \times 5)^2 \times 0.3 = 188.50 \text{ m/s}^2$$
• c) Total mechanical energy: $$E=\frac{1}{2}KA^2= \frac{1}{2}m\omega^2A^2= 13.88 \text{ J}$$

4. Block-spring system:

• a) Amplitude: $$0.6 \text{ m}$$
• b) Time period: $$T=2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{4}{20}} \approx 1.88 \text{ s}$$
• c) Maximum velocity: $$\omega A = 2\sqrt{5} \times 0.6 \approx 1.69 \text{ m/s}$$

Note: These are just sample calculations for better understanding. Numerical values may vary based on the actual problem statements and given parameters.