Question

For an SHM, graph between velocity (v) and displacement (x), in general is
Select one:
O a. Straight line
O b. Parabolic
O c. Hyperbolic
O d. Elliptical​

Answer 1

Answer:

$\boxed{\mathfrak{d. \ Elliptical}}$

Explanation:

Simple pendulum executes SHM for small displacements

$\rm x = Asin \omega t$

Velocity:

$\rm \implies v = \dfrac{dx}{dt} \\ \\ \rm \implies v = \dfrac{d}{dt} (Asin \omega t) \\ \\ \rm \implies v = A \omega cos \omega t$

Squaring the displacement equation:

$\rm \implies x ^{2} = A ^{2} sin ^{2} \omega t \\ \\ \rm sin ^{2} \omega t = 1 - cos ^{2} \omega t \\ \\ \rm \implies x ^{2} = A ^{2} (1 - cos ^{2} \omega t ) \\ \\ \rm \implies x ^{2} = A ^{2}- A ^{2}cos ^{2} \omega t \\ \\ \rm \implies A ^{2}cos ^{2} \omega t = A ^{2} - {x}^{2} \\ \\ \rm \implies cos ^{2} \omega t = \dfrac{ A ^{2} - {x}^{2} }{A ^{2}} \\ \\ \rm \implies cos ^{2} \omega t = 1 - \dfrac{ {x}^{2} }{A ^{2}} \\ \\ \rm \implies cos \omega t = \sqrt{ 1 - \dfrac{ {x}^{2} }{A ^{2}}}$

So,

$\rm \implies v = A \omega \sqrt{1 - \dfrac{ {x}^{2} } {A ^{2} } } \\ \\ \rm \implies v = A \omega \sqrt{ \dfrac{A ^{2} - {x}^{2} } {A ^{2} } } \\ \\ \rm \implies v = \cancel{A} \omega \dfrac{ \sqrt{ A ^{2} - {x}^{2}} } { \cancel{A } } \\ \\ \rm \implies v = \omega \sqrt{ A ^{2} - {x}^{2}} \\ \\ \rm \implies v ^{2} = \omega^{2} ( A ^{2} - {x}^{2}) \\ \\ \rm \implies \dfrac{ v ^{2} }{\omega^{2}} = ( A ^{2} - {x}^{2}) \\ \\ \rm \implies {x}^{2} + \dfrac{ v ^{2} }{\omega^{2}} = A ^{2} \\ \\ \rm \implies \dfrac{{x}^{2}}{A ^{2}} + \dfrac{ v ^{2} }{A ^{2}\omega^{2}} = \frac{ A ^{2}}{A ^{2}} \\ \\ \rm \implies \dfrac{{x}^{2}}{A ^{2}} + \dfrac{ v ^{2} }{A ^{2}\omega^{2}} = 1$

This equation represents ellipse.

Answer 2

Answer:

x=Asinωt or sinωt=x/A

v=dx/dt=Aωcosωt or cosωt=v/Aω

Thus, using sin

2

+cos

2

=1, x

2

/A

2

+x

2

/A

2

ω

2

=1, which gives the equation of ellipse.

Thus statement 1 is correct. Also, statement 2 is correct but it doesn't explain the reason