When a 0.20-kg block is suspended from a vertically hanging spring, it stretches

Question

When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.050 m to 0.060 m. The same block is attached to the same spring and placed on a horizontal, frictionless surface as shown. The block is then pulled so that the spring stretches to a total length of 0.10 m. The block is released at time t = 0 s and undergoes simple harmonic motion. What is the maximum acceleration of the block?

I know that the answer is 290 m/s^2 but i keep getting 98 every time i work it out! HELPPP!

Explanation / Answer

The mass of the block is 0.20 kg. When you hang it from the spring, gravitational force on the block is mg.

This extends the spring by x = 0.060 - 0.050 = 0.010 m

Therefore mg = kx.

---> k = 0.20 x 9.8/0.010 = 196 N/m. This is the value of the spring constant.

When it is then kept on a horizontal surface and stretched to a "total length of 0.10m" , this means x = 0.050m. i.e The amplitude of the oscillations is 0.05m

A simple harmonic oscilation can be represented by th equation x = A sin(t + ). A is the amplitude, is the angular frequency and is the initial phase.

"x" is the position of the block with respect to the mean position. i.e w.r.t to the natural state of spring

The force acting on the block due to the spring is kx, this will be maximum only if x is maximum, i.e x = A.

therefore kA = ma ---> maximum acceleration is 196 x 0.050 / 0.020 = 490 m/s2

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