Question

Two springs with negligible masses and force constants $k_1 = 200$ N/m and $k_2 = 160$ N/m are attached to the block of mass $m = 10$ kg as shown in figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time $t = 0$, sharp impulse of 50 N-s is given to the block with a hammer along the spring. Then the amplitude of spring will be K metre. Find 6K

Answer 1

5

Answer 2

Solution:

1. Effective Spring Constant:
   For springs in parallel,

   $$k_{\text{eff}} = k_1 + k_2 = 200 + 160 = 360\, \text{N/m}$$

2. Initial Velocity:
   An impulse $I = 50\, \text{N-s}$ gives the block a velocity:

   $$v = \frac{I}{m} = \frac{50}{10} = 5\, \text{m/s}$$

3. Amplitude $K$:
   At maximum displacement, all kinetic energy converts into spring potential energy:

   $$\frac{1}{2} m v^2 = \frac{1}{2} k_{\text{eff}} K^2$$

   Substituting values:

   $$\frac{1}{2} \times 10 \times 5^2 = \frac{1}{2} \times 360 \times K^2$$ $$125 = 180\, K^2 \quad \Rightarrow \quad K^2 = \frac{125}{180} = \frac{25}{36}$$ $$K = \frac{5}{6}\, \text{m}$$

4. Finding $6K$:

   $$6K = 6 \times \frac{5}{6} = 5\, \text{m}$$