Question

A block of mass 0.50 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.10 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.

Answer 1

T = 0.314 seconds.

Mass = 0.5 kg.

Using the Formula,

T = 2π/ω

∴ ω = 2π/T = 2 × 3.14/0.314

∴ ω = 20 J.

Now,  ω² = k/m

∴ k = mω²

∴ k = 0.5 × 400

∴ k = 200 N/m.

Now, Force = -kx  

∴ F = - 200 × 0.1

F = -20 N.

Hope it helps .

Answer 2

$<i>$ m = 0.50 kg,

$<b>A = 0.10 m,</b>$

$<b>T = 0.314 s</b>$

If the spring constant is k,

k=2π/T

=2π/0.314

=2*3.14/0.314

=20 s⁻¹

The magnitude of the maximum force =mk²A

=0.50*20²*0.10 N

=0.50*400*0.10 N

=20 N

or alternately,

T = 2π√(m/k)

→0.314 = 2π√(0.50/k)

→0.50/k = 0.314*0.314/4π²

→k = 0.50*4*π²/0.314*0.314 =200 N/m

The maximum force is at the extreme

F=k*A =200*0.10 N =20 N

When the spring is at rest with the block hanging (The mean position) the force by the spring on the block is

mg = 0.50*10 =5 N.

From this position the spring is stretched to the amplitude 0.10 m for which the extra force 20 N.

$<b>$ So the total force by the spring on the block is 20+5 = 25 N.