Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen

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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

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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

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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 = $\frac{\pi }{2}+wt-wt$ = $\frac{\pi }{2}$

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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.

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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

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As we know equation of SHM is x= Asinwt

V= dx/dt=Awsinwt

V_max=Awcoswt_max

= Aw

A=dv/dt=-wAwsinwt

= -w²Asinwt

A_max=-w²A

From above equations

$\frac{V_{max}}{A_{max}}$ =wA/w²A=1/w

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The bob is displaced through some angle

The restoring force $\tau =-mgsin\theta$ if $sin\theta$ is small then it is $\theta$ only.

$\tau \approx -mg\theta$

So torque is directly proportional to angle.

So it clear from the above equation that its period will be harmonic

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Acceleration is directly proportional to displacement.

The direction of acceleration is always towards the mean position that is opposite to displacement.

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Consider the diagram in which the block is displaced right through x

The right spring gets compressed by x developing a restoring force kx towards left on the block. The left spring is stretched by an amount x developing a restoring force kx towards left on the block

Hence total force =kx+kx= 2kx towards left.

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(i) In SHM y-t graph, zero displacement values correspond to mean position where velocity of the oscillator is maximum.

Where the crest and troughs represents extreme positions where displacement is maximum and velocity of the oscillator is minimum and is zero. Hence A

(ii) And also speed is maximum at mean position represented by B