Question

A massless spring with spring constant 16.4 N/m hangs vertically. A body of mass 0.193 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, and the (b) frequency and (c) amplitude of the resulting SHM.

Answer 1

Answer:

(A) 0.2306 m

(B) 1.467 Hz

(C) 0.1152 m

Explanation:

spring constant (K) = 16.4 N/m

mass (m) = 0.193 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) force = Kx,  where x = extension

   mg = Kx

   0.193 x 9.8 = 16.4x

   x = 0.1153 m

  now the mass actually falls two times this value before it gets to its equilibrium position ( turning  point ) and oscillates about this point

therefore

2x = 0.2306 m

(B) frequency (f) = \frac{1}{2π} x $\sqrt{\frac{k}{m}}$

     frequency (f) = \frac{1}{2π} x $\sqrt{\frac{16.4}{0.193}}$

     frequency = 1.467 Hz  

(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring

= \frac{0.2306}{2} =  0.1152 m