Question

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.160m . The maximum speed of the block is 3.95m/s . What is the maximum magnitude of the acceleration of the block?

Answer 1

Answer:

97.5 m/s^2

Explanation:

The magnitude of the maximum speed in a SHM is given by

$v_{max} = \omega A$

where

$\omega$ is the angular frequency of the motion

A is the amplitude

We can rewrite the equation as

$\omega=\frac{v_{max}}{A}$ (1)

The magnitude of the maximum acceleration is given instead by

$a_{max} = A \omega^2$

And using (1), we can rewrite it as

$a_{max}=A (\frac{v_{max}}{A})^2 = \frac{v_{max}^2}{A}$

In this problem, we have

A = 0.160 m (amplitude of the motion)

$v_{max}=3.95 m/s$ (maximum speed)

Therefore, using the above equation we find the magnitude of the maximum acceleration:

$a_{max}=\frac{(3.95 m/s)^2}{0.160 m}=97.5 m/s^2$

Answer 2

The maximum magnitude of the acceleration of the block will be

$a_m=97.5 \dfrac{m}{s^2}$

What will be the maximum magnitude of the acceleration of the block?

In an SHM the formula for maximum acceleration is given by

$V_{max}=wA$

Here

$w$ is the angular frequency of the motion

A is the amplitude

We can rewrite the equation as

$w=\dfrac{V_{max}}{A}$

The magnitude of the maximum acceleration is given instead by

$a_{max}=Aw^2$

By putting the value of w from the above equation

$a_{max}=A(\dfrac{V_{max}}{A} )^2$

$a_{max}=\dfrac{V_{max}^2}{A}$

By putting the value of

$A=0.160m$

$V_{max}=3.95\dfrac{m}{s}$

$a_{max}=\dfrac{3.95^2}{0.160} =97.5\dfrac{m}{s^2}$

Thus  the maximum magnitude of the acceleration of the block will be

$a_m=97.5 \dfrac{m}{s^2}$

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