A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(−25𝑥)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ______ cm.

Question

A 1.5 kg block is placed on a frictionless surface and attached to a spring with a spring constant of 100 N/m. If the block is displaced by 0.2 m from its equilibrium position, what is the potential energy stored in the spring?

Answer 1

To solve this problem, we need to find the amplitude of the particle's simple harmonic motion (SHM) given the periodic force, mass, and maximum speed.

Step 1: Understanding the Force

The force acting on the particle is F = -25x. This is a restoring force characteristic of simple harmonic motion, often given by F = -kx, where k is the spring constant. Thus, k = 25 N/m.

Step 2: Calculating Angular Frequency

The angular frequency ω can be found using the formula ω = √(k/m). The mass m = 250 g = 0.25 kg. Therefore, ω = √(25/0.25) = √100 = 10 rad/s.

Step 3: Using Maximum Speed

The maximum speed v_max in SHM is given by v_max = ωA, where A is the amplitude. Here, v_max = 4 m/s. Solving for A, we get A = v_max/ω = 4/10 = 0.4 m = 40 cm.

Step 4: Verification within the Given Range

The computed amplitude A = 40 cm, which falls within the provided range of 40 to 40, confirming the correctness of the solution.

Conclusion

The amplitude of the motion is 40 cm.

A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(−25𝑥)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ______ cm.

Correct Answer: 40

Solution and Explanation

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