At a given point of time the value of displacement of a simple harmonic oscillator is given as y=A cos (30°). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is 1.0 × 10ˣ Nm⁻¹. The value of x is ____________.

Solution

Physics sir · 4 steps

Kinetic energy is total energy minus potential energy. Find how much of it is left at y = A cos 30°.

1. Energy at position y. Total energy in SHM is 1/2kA² and potential energy at displacement y is 1/2ky². Kinetic energy is what remains.

   K = 1/2k(A² - y²)

2. Use the displacement. With y = A cos 30° = √3/2A, you get y² = 3/4A². So A² - y² = 1/4A² and K = 1/8kA².
3. Put in numbers. Take A = 0.4 m, so A² = 0.16 m². Then 200 = (k × 0.16)/8 = 0.02k, which gives k = 10⁴ N m⁻¹.
4. So. k = 1.0 × 10⁴ N m⁻¹, so x = 4.

The common slip: 0. Using the amplitude as 40 instead of 0.4 m gives A² = 1600 and k = 1 N m⁻¹, so x = 0, not 4.

The idea to keep. In SHM, K = 1/2k(A² - y²) at any displacement y. Write y as a fraction of A first, and always put lengths in metres before you use k in N m⁻¹.

Answer from the NTA final answer key. Solution by PrepWiser.