Class 11th

  $tan\alpha =\frac{v_{max}}{t_1}=a_1$

$tan\beta =\frac{v_{max}}{t_2}=a_2$

$\Rightarrow \frac{a_1}{a_2}=\frac{t_2}{t_1}\Rightarrow \frac{t_1}{t_2}=\frac{a_2}{a_1}$

Kindly go through the solution

-> $v=\omega \sqrt[]{A^2-x^2}\Rightarrow \frac{v^2}{A^2{\omega }^2}+\frac{x^2}{A^2}=1$ Path is ellipse.

Let the spring is in extension state and   $a_B>a_A,{\overrightarrow{a}}_{AB}=-a_A\widehat{i}-\left(-a_B\widehat{i}\right)=\left(-a_A+a_B\right)\widehat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B           . (1)

kx = Ma                  . (2)

$\Rightarrow a_B=\frac{F}{M}-a$

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

Stress = Y * strain

-> $\frac{T_1}{A}=Y*-\frac{\left({\mathcal{l}}_1-\mathcal{l}\right)}{\mathcal{l}}\left(i\right)$  

  $\frac{T_2}{A}=Y*\frac{\left({\mathcal{l}}_2-\mathcal{l}\right)}{\mathcal{l}}\left(ii\right)$                          

$\Rightarrow$ $\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}$                      

$\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_2}{T_1-T_2}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$            

$y=\alpha x-\beta x^2$

$\Rightarrow \frac{dy}{dx}=\alpha -2\beta x=0$  

$\Rightarrow x=\frac{\alpha }{2\beta }$           

$y_{max}=\alpha *\frac{\alpha }{2\beta }-\beta *{\left(\frac{\alpha }{2\beta }\right)}^2=\frac{{\alpha }^2}{4\beta }$           

Range = $2x=\frac{\alpha }{\beta }=\frac{2u^{2}sin\theta .cos\theta }{g}$  

On comparing with

y = x tan θ - $\frac{g{x}^2}{2u^{2}co{s}^2\theta }$  

mg – T = ma       . (1)

T * R = l a               . (2)

a = a R                    . (3)

With the help of equations (1), (2) and (3), we get

-> $a=\frac{mg}{m+\frac{l}{R^2}}$  

$v=\sqrt[]{2ah}=\omega R$  

$\Rightarrow {\omega }^2=\frac{2mgh}{l+m{R}^2}$     

You can read the classification of elements and periodicity in properties class 11 notes to study class 11 ch3 and score well in class 11th exams and NEET, and JEE exams.

When the elements having similar properties repeat at regular intervals, it is called periodicity in the properties of elements.