Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen: Overview, Questions, Preparation

Explanation-a) the weight of the body changes during oscillations.

b)considering the situations in two extreme positions

we can say mg-N= ma

so at highest point the platform is accelerating downward.

N=mg-ma

a=w²A

N=mg-mw²A

A= amplitude of motion m=50kg v=2m/s

w=2 $\pi v=4\pi rad/s$

 A= 5cm = 5 $\times {10}^{-2}m$

N= 50 $\times 9.8-50\times {\left(4\pi \right)}^2\times 5\times {10}^{-2}=95.5N$

When it is accelerating towards mean position that is vertically upwards

N-mg=ma=Mw²A

N=mg+mw²A

N=m(g+w²A)

N= 50[9.8+ ${\left(4\pi \right)}^2\times 5\times {10}^{-2}$ ]

N= 884N

Machine reads the normal reaction

Maximum weight =884N

Minimum weight=95.5N

2. A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation?

Explanation- when the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x’ is the extension in that case

F= kx’

F=mg

Mg=kx’

X’=mg/k

By dividing x by x’

x/x’= $\frac{2mg/k}{mg/k}=2$

so x=2x’

x’=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x’=2cm

b) time period of the oscillating system depends on mass spring constant given by

T= $2\pi \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $\times {10}^{-2}m$

m/k =4 $\times {10}^{-2}$ /2g

k/m=g/2 $\times {10}^{-2}$

$v=\frac{1}{T}\sqrt[]{\frac{k}{m}}$

$v=\frac{1}{2\times 3.14}\sqrt[]{\frac{g}{2\times {10}^{-2}}}$

= $\frac{10}{628}\times 2.21$ =3.51Hz

3. A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

Explanation- let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $\rho A{x}_{o}g$

When it is displaced by further displacement x, the buoyant force is A(x_o+x) $\rho g$

Net restoring force = buoyant forces -weight

=A(x_o+x) $\rho g$ -mg

=A $\rho gx$

As displacement x is downward and restoring force is upward

F_{restoring =}-A $\rho gx$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

T= 2 $\pi \sqrt[]{\frac{m}{k}}=2\pi \sqrt[]{\frac{m}{A\rho g}}$

4. One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Explanation-let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $\rho =$ density of the liquid

PE= dmgx=A $\rho dxgx$

So total PE of the column

= ${\int }_0^{h_1}Adxg\rho x$ = $Ag\rho {\int }_o^{h_1}xdx=A\rho g\frac{h_1^2}{2}$

But h₁=lsin45

PE=A $\rho$ gl²sin²45/2

Similarly PE of right column = A $\rho$ gl²sin²45/2

Total PE = A $\rho$ gl²sin²45/2+ A $\rho$ gl²sin²45/2

= A $\rho$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A\rho g}{2}[{\left(l-y\right)}^2+{\left(l+y\right)}^2-l^2$ ]

= $\frac{A\rho g}{2}[2{(l}^2+y^2)]$ =A $\rho g{(l}^2+y^2)$

Change in KE = ½ mv²

m=A $\rho 2l$

change in KE= 1/2A $\rho 2l{v}^2$ =A $\rho l{v}^2$

so from above eqn

change in PE +change in KE = A $\rho g{(l}^2+y^2)$ +A $\rho l{v}^2$

system being conservative

so change in total energy =0

A $\rho g{(l}^2+y^2)+A\rho l{v}^2$ =0

Differentiating both with respect to time t

A $\rho g\left(2yv\right)+A\rho l\left(2v\right)a=0$ =0

Dy/dt =v and dv/dt=a

A

(gy+la)2A $\rho v$ =0

2A $\rho v=constant$ and 2A $\rho v\ne 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $\pi \sqrt[]{\frac{l}{g}}$

 

5. A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Explanation- the gravitational force on the particle at a distance r from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the positiin of the particles . the external shells exert no force on the particle.

Let g’ be the acceleration at P

So g’ =g(1-d/R)=g(R-d/R)

R-d=y

g’=gy/R’

F=-mg’= -mgy/R

F $\propto -y$

Ma=-Mgy/R , a = -gy/R

Comparing a=-w²y

w²=g/R

T=2 $\pi \sqrt[]{\frac{R}{g}}$

7.A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Fig) The amplitude is θ o . The string snaps at θ = θ₀ /2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θ₀ to be small so that sin _{o 0} and cos ₀

Explanation- let us assume t=0 when $\theta =\theta$ _o then $\theta$ = $\theta$ ₀coswt

Given a seconds pendulum w=2 $\pi$

$\theta$ = ${\theta }_{o}cos2\pi t$

At time t, let $\theta ={\theta }_o/2$

Cos2 $\pi t_1=\frac{1}{2}$  , t₁=1/6

$\frac{d\theta }{dt}$ =-( ${\theta }_{o}2\pi$ )sin2 $\pi t$

t=t₁=1/6

d $\frac{\theta }{dt}$ =- ${\theta }_{o}2\pi$ sin $\frac{2\pi }{6}=-\sqrt[]{3}\pi {\theta }_o$

the linear velocity is u=- $\sqrt[]{3}\pi {\theta }_{o}l$

the vertical component is U_y= $-\sqrt[]{3}\pi lsin\left(\frac{{\theta }_o}{2}\right)$

the horizontal component U_x=- $-\sqrt[]{3}\pi lcos\left(\frac{{\theta }_o}{2}\right)$

at the time it snaps the vertical height is

H’=H+l(1-cos $\left(\frac{{\theta }_o}{2}\right)$ )

Let the time require for fall be t , then

H’= H+l (1 $-cos\left(\frac{{\theta }_o}{2}\right)$ )

Let the time required for fall be at t then

H’=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3}\pi {\theta }_{o}lsin\frac{{\theta }_o}{2}t-H^{\text{'}}=0$

t= $\frac{-\sqrt[]{3}\pi {\theta }_{0}sin\frac{{\theta }_o}{2}\pm \sqrt[]{3{\pi }^2{\theta }_o^2{sin}^2\frac{{\theta }_o}{2}+2gH\text{'}}}{g}$

given that ${\theta }_o$ is small , hence neglecting terms of order ${\theta }_o^2$ and higher

t= $\sqrt[]{\frac{2H^{\text{'}}}{g}}$

H’ $\approx H+l(1-1)$

t= $\sqrt[]{\frac{2H}{g}}$

the distance travelled in the x direction is u_xt to the left of where the bob is snapped 

X= Uxt= $\sqrt[]{3}\pi {\theta }_{o}lcos\left(\frac{{\theta }_o}{2}\right)\sqrt[]{\frac{2H}{g}}s$

cos $\frac{{\theta }_o}{2}\approx 1$

X= $\sqrt[]{3}\pi {\theta }_{o}l\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

At the time of snapping , the bob was at a horizontal distance of $lsin\frac{{\theta }_o}{2}\approx l\frac{{\theta }_0}{2}$ from A

Thus, the distance of bob from A where it meets the ground is

= $\frac{l{\theta }_o}{2}-X=\frac{l{\theta }_o}{2}-\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi \right)$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}\right)$