Oscillations

Time period of second pendulum is 2 seconds.

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

m = 4kg

U = 4 (1 – cos 4x) J

$F=-\frac{\partial \upsilon }{\partial x}=-4*4sin4x$

$\therefore a=-\frac{16}{4}sin4x=-4sin4x=-16x$

$\Rightarrow a=-16x$

${\omega }^2=16\Rightarrow \omega =4=\frac{2\pi }{T}\Rightarrow T=\frac{2\pi }{4}=\frac{\pi }{2}sec$

k = 2

A = 8 cm

T = 6 sec

q = 60° from A to B

During reaching the point its maximum amplitude from point (A)

$\theta =60°=\frac{{\pi }^C}{3}$

$\theta =\omega t$                

$\frac{\pi }{3}=\frac{2\pi }{6}t$                              

t = 1 sec

$x=sin\pi \left(t+\frac{1}{3}\right)$

$v=\frac{dx}{dt}=cos\left[\pi \left(t+\frac{1}{3}\right)\right]*\pi$

$=cos\left(\pi +\frac{\pi }{3}\right)*\pi$

$=-cos\frac{\pi }{3}*\left(\pi \right)$

$=-\frac{\pi }{2}=-1.57m/sec$              

Speed = 157 cm/sec

This is a multiple choice answer as classified in NCERT Exemplar

Time period of simple pendulum T=2s

For simple pendulum T= $2\pi \sqrt[]{\frac{l}{g}}$  where l is length and g = acceleration due to gravity.

T_e=2 $\pi \sqrt[]{\frac{l_e}{g_e}}$

On the surface of the moon T_m= 2 $\pi \sqrt[]{\frac{l_m}{g_m}}$

$\frac{T_e}{T_m}$ = $\frac{2\pi }{2\pi }\sqrt[]{\frac{l_e}{g_e}}*\sqrt[]{\frac{g_m}{l_m}}$

T_e=T_m to maintain the second's pendulum time period

1= $\sqrt[]{\frac{l_e}{g_e}}*\sqrt[]{\frac{g_m}{l_m}}$ …………….1

But the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth,

g_m= $\frac{g_e}{6}$

squaring equation 1 and putting this value

1= $\frac{l_e}{l_m}*\frac{g_e/6}{g_e}=\frac{l_e}{l_m}*\frac{1}{6}$

l_m=1/6l_e = 1/6 m

This is a multiple choice answer as classified in NCERT Exemplar

Potential energy of a simple harmonic oscillator is = ½ kx²=1/2mw²x²

K=mw²

When x=0 PE=0

When x= $?A$ , PE=maximum

=1/2 mw²A²

KE of a simple harmonic oscillator =1/2 mv²

= 1/2 m [w $\sqrt[]{A^2-x^2}$ ] ²

= ½ mw² (A²-x²)

This is also parabola if plot KE against displacement x

KE= 0 at x= $?A$

KE=1/2mw²A² at x=0

Now total energy of the simple harmonic oscillator =PE+KE

= ½ mw²x²+1/2mw² (A²-x²)

TE= ½ mw²A²

So the curve according to that is

This is a multiple choice answer as classified in NCERT Exemplar

As we know x= acoswt

V =dx/dt= a (-sinwt)w=-wasinwt

V=-wasinwt

= wacos ( $\frac{\pi }{2}+wt$ )

Phase of velocity = $\frac{\pi }{2}+wt$

So difference in phse of velocity to that of phase of displacement = $\frac{\pi }{2}+wt-wt$ = $\frac{\pi }{2}$

This is a multiple choice answer as classified in NCERT Exemplar

As the particle on reference circle moves in anticlockwise direction. The projection will move from P to O towards left.

Hence in the position shown the velocity is directed from P' to P'' i.e from right to left . hence sign is negative.

This is a multiple choice answer as classified in NCERT Exemplar

In the diagram

the motion of a particle  executing SHM between A and B

Total distance travelled while it goes from A to B and returns to A is=AO+OB+BO+OA

= A+A+A+A=4A

So ratio of distance and amplitude =4A/A=4