A body weighing 1 kg first vibrates on a thread, and then on a spring with a stiffness of 5N / m.

A body weighing 1 kg first vibrates on a thread, and then on a spring with a stiffness of 5N / m. At what length of the thread will the vibration frequencies received by the pendulums be the same?

To determine the required length of the thread of the obtained mathematical pendulum, we use the equality: 1/2 * п * √ (k / m) = ν (vibration frequency) = 1/2 * п * √ (g / l), from where we express: l = g * m / k.

Constants and variables: g – gravitational acceleration (g ≈ 10 m / s2); m – body weight (m = 1 kg); k is the stiffness of the spring (k = 5 N / m).

Calculation: l = g * m / k = 10 * 1/5 = 2 m.

Answer: Identical vibration frequencies will be obtained when the length of the thread of the obtained mathematical pendulum is 2 meters.