Question

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 given an initial velocity of 3.5 m/s starting at the equilibrium position, what is the maximum displacement of the block from its equilibrium position

Answer 1

Answer:

The maximum displacement ,A= 0.047 m

Explanation:

Given that

Mass, m = 45 g= 0.045 kg

spring constant ,K= 240 N/m

Initial velocity at equilibrium ,v= 3.5 m/s

Lets take the maximum displacement is A

As we know that when the mass reached at the extreme position then the velocity of the mass will become zero.

From energy conservation

$\dfrac{1}{2}KA^2=\dfrac{1}{2}mv^2$

$KA^2=mv^2$

$A=\sqrt{\dfrac{mv^2}{K}}$

Now by putting the values

$A=\sqrt{\dfrac{0.045\times 3.5^2}{240}}$

A=0.047 m

The maximum displacement ,A= 0.047 m

Answer 2

Answer:

x = 4.79 cm

Explanation:

given,

mass of block = 45 g

spring constant = 240 N/m

initial velocity = 3.5 m/s

maximum displacement = ?

using conservation of energy

loss of KE = grain in spring constant

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2$

$mv^2 =kx^2$

$x = \sqrt{\dfrac{mv^2}{k}}$

$x = \sqrt{\dfrac{0.045\times 3.5^2}{240}}$

x = √0.00229

x = 0.0479 m

x = 4.79 cm