Simple Harmonic Motion Concept Page - 7

Definition

Problem on Spring in horizontal motion

Example:
The spring shown in figure is unstretched when a man starts pulling the block. The mass of the block is M$M$. If the man exerts a constant force F$F$.
Find the amplitude of the motion of the block.
Solution:The block will oscillate about the position
where F=−kx$F=-kx$
⇒Fk$\Rightarrow {\displaystyle \frac{F}{k}}$ is amplitude

Example

TIme period of spring mass system

Example: A mass M$M$ is suspended from a light spring. An additional mass m$m$ added to it displaces the spring further by a distance x$x$, then find its time period.

Solution:
T=2πmk−−−√$T=2\pi \sqrt{{\displaystyle \frac{m}{k}}}$
kx=mg⇒1k=xmg$kx=mg\Rightarrow {\displaystyle \frac{1}{k}}={\displaystyle \frac{x}{mg}}$
When extra mass is added, total mass =(M+m)$=(M+m)$
∴T=2π(M+m)k−−−−−−−−√=2πx(M+m)mg−−−−−−−−−√$\therefore T=2\pi \sqrt{{\displaystyle \frac{(M+m)}{k}}}=2\pi \sqrt{{\displaystyle \frac{x(M+m)}{mg}}}$

Example

Application of conservation of energy for spring in vertical plane

Example:
A toy gun consists of a spring and a rubber dart of mass 25$25$ gm. When the spring is compressed by 4$4$ cm and the dart is fired vertically, it projects the dart to a height of 2$2$ m. If the spring is compressed by 8$8$ cm and the same dart is projected vertically, the dart will rise to a height of?
Solution:12Kx2=mgh${\displaystyle \frac{1}{2}}K{x}^2=mgh$ 
hence , x2∝h$x^2\propto h$ 
hence , h2=8242${\displaystyle \frac{h}{2}}={\displaystyle \frac{8^2}{4^2}}$
∴h=8m$\therefore h=8m$

Example

Write the force equation for a mass attached to a spring in vertical motion

Example: A mass m$m$ is undergoing SHM in the vertical direction about the mean position y0$y_0$ with amplitude A and angular frequency ω$\omega$. At a distance y$y$ from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m$m$. Find the distance y$y$ (measured from the mean position) such that the height h$h$ attained by the block is maximum.
Solution:
Let   v$v$  be the upward velocity of the block when it detaches from the spring at a distance y$y$ above the mean position.v=wA2−y2−−−−−−√⟹v2=w2(A2−y2)$v=w\sqrt{A^2-y^2}⟹v^2=w^2(A^2-y^2)$       ..........(1)
At the detaching point, the block will continue to move upwards due to inertia and reach the  height  h$h$ above the mean position. At that height the velocity of the block will be zero.
Thus        0−v2=2(−g)s⟹s=v22g$0-v^2=2(-g)s⟹s={\displaystyle \frac{v^2}{2g}}$
Now the total height     h=s+y=v22g+y$h=s+y={\displaystyle \frac{v^2}{2g}}+y$   
∴h=w2(A2−y2)2g+y$\therefore h={\displaystyle \frac{w^2(A^2-y^2)}{2g}}+y$
For  h$h$  to be maximum,     dhdy=0${\displaystyle \frac{dh}{dy}}=0$
Thus       w22g(−2y)+1=0${\displaystyle \frac{w^2}{2g}}(-2y)+1=0$         ⟹y=gw2$⟹y={\displaystyle \frac{g}{w^2}}$

Definition

Find equilibrium position of a mass attached to a spring in different arrangements

Example: Two solid cylinders connected with a short light rod about common axis have radius R$R$ and total mass M$M$ rest on a horizontal table top connected to a spring of spring constant k$k$ as shown. The cylinders are pulled to the left by x$x$ and released. There is sufficient friction for the cylinders to roll. Find time period of oscillation.

Solution:
Here, τ=Iα=−kx.R$\tau =I\alpha =-kx.R$
or (MR22+MR2)α=−kxR$({\displaystyle \frac{M{R}^2}{2}}+M{R}^2)\alpha =-kxR$
or Rα=−2k3Mx$R\alpha =-{\displaystyle \frac{2k}{3M}}x$
now, a=Rα=−2k3Mx$a=R\alpha =-{\displaystyle \frac{2k}{3M}}x$
thus, ω=2k3M−−−−√$\omega =\sqrt{{\displaystyle \frac{2k}{3M}}}$
Time period T=2πω=2π3M2k−−−−√$T={\displaystyle \frac{2\pi }{\omega }}=2\pi \sqrt{{\displaystyle \frac{3M}{2k}}}$

Formula

Write force equations of an arrangement of springs in series and calculate equivalent spring constant

For the given arrangement of springs and mass for an extension x$x$ of the spring:
F=k1x+k2x$F=k_{1}x+k_{2}x$
for equivalent spring constant k.
kx=k1x+k2x$kx=k_{1}x+k_{2}x$
k=k1+k2$k=k_1+k_2$

Example

Solve problems on describing motion of two blocks attached with a spring

Example: A block of mass m=1kg$m=1kg$ placed on top of another block of mass M=5kg$M=5kg$ is attached to a horizontal spring of force constant k=20N/m$k=20N/m$ as shown in figure. The coefficient of friction between the blocks is μ$\mu$ where as the lower block slides on a friction less surface. The amplitude oscillation is 0.4 m. What is the minimum value of μ$\mu$ such that the upper block does not slip over the lower block?
The upper block does not slip over the lower block when the restoring force is balanced by the friction force of lower block against ground. i.e, kx0=μ(M+m)g$k{x}_0=\mu (M+m)g$
or μ=kx0(M+m)g=20×0.4(5+1)10=0.133$\mu ={\displaystyle \frac{k{x}_0}{(M+m)g}}={\displaystyle \frac{20\times 0.4}{(5+1)10}}=0.133$

Example

Solve problems in which part of the motion is SHM

Wk=x1+x2=Wk1+Wk2${\displaystyle \frac{W}{k}}=x_1+x_2={\displaystyle \frac{W}{k_1}}+{\displaystyle \frac{W}{k_2}}$

Equivalent spring constant: k=k1k2k1+k2$k={\displaystyle \frac{k_1{k}_2}{k_1+k_2}}$

Example

SHM in spring in accelerated frame of reference

Example:
A block of mass 1$1$ kg is kept on smooth floor of a truck. One end of a spring of force constant 100$100$ N/m is attached to the block and other end is attached to the body of truck as shown in the figure. At t=0$t=0$, truck begins to move with constant acceleration 2$2$ m/s2${}^2$. Find the amplitude of oscillation of block relative to the floor of truck.
Solution:Let x0$x_0$ is the compression in equilibrium. Then
kx0=ma$k{x}_0=ma$
x0=mak${\displaystyle x_0=\frac{ma}{k}}$
   =1×2100${\displaystyle =\frac{1\times 2}{100}}$
   =0.02$=0.02$ m
Amplitude=x0$=x_0$
   =0.02$=0.02$ m

Example

Problem of various combination of Springs

Let the spring constant of the spring be K$K$.
Thus the effective spring constant of the parallel combination as shown in the figure=2K$2K$
Thus in a series combination of above springs, effective spring constant=Ki=K×2KK+2K$K_i={\displaystyle \frac{K\times 2K}{K+2K}}$=2K3$={\displaystyle \frac{2K}{3}}$
In parallel combination of above springs, effective spring constant=Kp=K+2K=3K$K_p=K+2K=3K$
Thus KiKp=29

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