Question

Show that for a wave travelling on a string
ymaxνmax=νmaxαmax,
where the symbols have usual meanings. Can we use componendo and dividendo taught in algebra to write
ymax+νmaxνmax-νmax=νmax+αmaxνmax-αmax?

Answer 1

The Given Relation is Incorrect!

Explanation:

For the required Answer, Let us take the case of velocity and acceleration separately,

Case I

We know that, $y = A(sin\omega t + kx)$

or $Y_m_a_x = A$

& $v = \frac {dy}{dt} = A\omega (cos\omegat -kx)$

$v_m_a_x = A\omega$

Now, for acceleration, Acceleration$(a) = \frac {dv}{dt}$

= $- A \omega^2 sin(\omega t -kx)$

= $\left | a \right |=A\omega^2$

$a_m_a_x = A\omega^2$

Now, $\frac {y_m_a_x}{v_m_a_x} = \frac {v_m_a_x}{a_m_a_x}$

$\frac {A}{A\omega} = \frac {A\omega}{A\omega^2}$

So, $\frac {1}{\omega} = \frac {1}{\omega}$

or LHS = RHS

The Given relation is dimensionally incorrect,

So it is not Possible.

Answer 2

Given that,

First relation is

$\dfrac{y_{max}}{v_{max}}=\dfrac{v_{max}}{a_{max}}$...(I)

Second relation is

$\dfrac{y_{max}+v_{max}}{y_{max}-v_{max}}=\dfrac{v_{max}+a_{max}}{v_{max}-a_{max}}$....(II)

We know that,

The equation of wave is

$y=A\sin(\omega t-kx)$

The maximum value of displacement is

$y_{max}=A$

The equation of velocity of wave is

$v=\dfrac{dy}{dt}=A\omega(\cos\omega t-kx)$

The maximum value of velocity is

$v_{max}=A\omega$

We need to calculate the acceleration of the wave

Using equation of velocity of wave

$v=A\omega(\cos\omega t-kx)$

$a=\dfrac{dv}{dt}$

Put the value into the formula

$a=\dfrac{d}{dt}(A\omega(\cos\omega t-kx))$

$a=A\omega^2$

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

We need to prove the given first relation

Using equation (I)

$\dfrac{y_{max}}{v_{max}}=\dfrac{v_{max}}{a_{max}}$

Put the value in to the formula

$\dfrac{A}{A\omega}=\dfrac{A\omega}{A\omega^2}$

$\dfrac{1}{\omega}=\dfrac{1}{\omega}$

L.H.S=R.H.S

(b). We need to prove the given second relation

Using given equation

$\dfrac{y_{max}+v_{max}}{y_{max}-v_{max}}=\dfrac{v_{max}+a_{max}}{v_{max}-a_{max}}$

According to given relation,

The given relation is dimensionally incorrect because we can not add the velocity in displacement nor add the acceleration in velocity.

So, This relation is not correct.

Hence, (a). First relation is proved.

(b). Second relation is not correct.