Question

A block is attached to a spring and set in motion on a horizontal frictionless surface by pulling the block back a distance 10cm from equilibrium. Now, replace the block with one double the mass and set the block into motion again by pulling the block a distance 10cm from equilibrium and releasing. How will the following new quantities relate to the quanities with the original block? (If it will be larger by a factor of 2, choose double. sqrt stands for square root.)

Amplitude

Maximum

Acceleration

Maximum Force of the Spring on the Block

Period of oscillation

Options are: quarter, half, sqrt(1/2) ,same, sqrt(2), double, quadruple

Answer 1

motion of a spring can be approximated as,

$x(t)=Acos(\omega t)$ ------------------(1)

where A is the amplitude and is equal to maximum displacement from equilibrium position,

A=10 cm

$\omega$ is the frequency of vibration of spring

x(t) denotes the position of block at any time t

now there will be a restoring force set up in spring, according to Hooke's law,

F=-kx

also, $F=m\frac{\mathrm{d^2}x }{\mathrm{d} t^2}$

combining both equations,

$-kx=m\frac{\mathrm{d^2}x }{\mathrm{d} t^2}$ ------------------(2)

since

$x(t)=Acos(\omega t)$

so $\frac{\mathrm{d}x }{\mathrm{d} t}=-A\omega sin(\omega t)$ -------------------(3)

and $\frac{\mathrm{d^2}x }{\mathrm{d} t^2}=-A\omega^2 cos(\omega t)$ -----------(4)

substituting (4) in (2),

$-kx=-Am\omega^2 cos(\omega t)=-\omega^2x$

$\omega=\sqrt{\frac{k}{m}}$ ----------------(5)

1. since amplitude is the maximum displacement and it is equal to 10 cm in both cases. So it remains same.

2. Acceleration, $a=\frac{\mathrm{d^2}x }{\mathrm{d} t^2}$

from equation (4), $a=\frac{\mathrm{d^2}x }{\mathrm{d} t^2}=-A\omega^2 cos(\omega t)$

acceleration will be maximum when $cos(\omega t)=1$

which means, $\omega t=90^{\circ}$

so maximum acceleration, $a_{max}=-A\omega^2$

using equation (5)

$a_{max}=-10*\frac{k}{m}$

on doubling the value of mass

$a'_{max}=-10*\frac{k}{2m}=a_{max}/2$

so maximum accleration reduces by half.

3. Force on block,

$F=m\frac{\mathrm{d^2}x }{\mathrm{d} t^2}$

$F=ma$

maximum force, $F_{max}=ma_{max}$

on doubling the value of mass, $F'_{max}=2ma'_{max}$

$F'_{max}=2m\frac{a_{max}}{2}=ma_{max}=F$

no change in force

so force remains same.

4. since time period, $T=\frac{2\pi }{\omega }$

so $T=2\pi \sqrt{\frac{m}{k}}$

on doubling mass,

$T'=2\pi \sqrt{\frac{2m}{k}}$

$T'=\sqrt{2}\left (2\pi \sqrt{\frac{m}{k}} \right )$

$T'=\sqrt{2}T$ so time period or period of oscillation changes by sqrt (2).