A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a frictionless surface. If the block is displaced 3.5 cm from its equilibrium position and released so that its initial velocity is zero, what is its maximum velocity

The maximum velocity of the block is 2.56 m/s.

Explanation:

Given;

mass of the block, m₁ = 45g = 0.045 kg

spring constant, k = 240 N/m

displacement of the block, x = 3.5 cm = 0.035 m

Apply the principle of conservation of energy;

P.E = K.E

¹/₂kx² = ¹/₂m(v-u)²

where;

v is the maximum velocity of the block

u is the initial velocity of the block, u = 0

¹/₂kx² = ¹/₂mv²

kx² = mv²

Therefore, the maximum velocity of the block is 2.56 m/s.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

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