NCERT Solutions for Class 11 Science Physics Chapter 14 - Oscillations

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Answer:

A simple harmonic motion is produced when a force (called restoring force) proportional to and in opposite direction to the displacement acts on a particle. All sine and cosine functions of time (t) are simple harmonic in nature. Hence the motion describe in the equation given in the question is in simple harmonic motion.

We know if the period of $f\left(x\right)\to T$, then the period of $f\left(ax\right)\to \frac{T}{a}$ and $f\left(\frac{x}{a}\right)\to aT$
Since the period of $\cos \left(t\right)\to 2\pi$, then the period of $\cos \left(2\omega t\right)\to \frac{2\pi }{2\omega }=\frac{\pi }{\omega }$
In the equation, $y=3 \cos \left(\frac{\pi }{4}-2\omega t\right)$, 3 is the amplitude and $\frac{\pi }{4}$ will just shift the graph hence 3 and$\frac{\pi }{4}$ won't have any effect on the period.

Hence the correct answer is option (b).

Answer:

We know $\sin \left(3\mathrm{t}\right)=3\sin \left(\mathrm{t}\right)-4{\sin }^3\left(\mathrm{t}\right)$

So, $4\mathrm{y}=3\sin \left(\mathrm{\omega t}\right)-\sin \left(3\mathrm{\omega t}\right)$
$$\begin{gathered} \Rightarrow 4\frac{\mathrm{dy}}{\mathrm{dt}}=3\mathrm{\omega } \mathrm{cos\omega t}-3\mathrm{\omega } \cos 3\mathrm{\omega t} \\ \Rightarrow 4\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-3{\mathrm{\omega }}^{2 }\mathrm{sin\omega t}+9{\mathrm{\omega }}^2 \mathrm{sin\omega t} \\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=\frac{-3{\mathrm{\omega }}^{2 }\mathrm{sin\omega t}+9{\mathrm{\omega }}^2 \mathrm{sin\omega t}}{4} \end{gathered}$$
We can see from the above expression $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}$ is not proportional to y.
Hence, motion is not SHM.

As the expression $4\mathrm{y}=3\sin \left(\mathrm{\omega t}\right)-\sin \left(3\mathrm{\omega t}\right)$ is the sum of 2 SHM's hence it will be periodic.

Hence the correct answer is option (b).

Answer:

 In simple harmonic motion, acceleration is proportional and opposite to displacement. Out of all four options, only a_x = –2x satisfies the definition of SHM.

Hence the correct answer is option (d).

Answer:

The motion of an oscillating liquid column in a U tube is SHM with the period, $2\pi \sqrt{\frac{l}{g}}$ ,where l is the height of liquid column in one arm of the U tube in the equilibrium position of liquid. Therefore, T is independent of the density of the liquid.
Hence the correct answer is option (c).

Answer:

$$\begin{gathered} \mathrm{x} = \mathrm{a} \cos \mathrm{\omega t} ...\left(1\right) \\ \mathrm{y} = \mathrm{a} \sin \mathrm{\omega t} ...\left(2\right) \\ \mathrm{Squaring} \mathrm{and} \mathrm{adding} \left(1\right) \mathrm{and} \left(2\right) \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \end{gathered}$$

The above equation represents a circle of radius a.

Hence the correct answer is option is (c).
 

Answer:

$\mathrm{y} = \mathrm{a} \sin \mathrm{\omega t} + \mathrm{b} \cos \mathrm{\omega t}$
$\mathrm{y} =\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \left(\frac{\mathrm{a} \sin \mathrm{\omega t} + \mathrm{b} \cos \mathrm{\omega t}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)$
Let $\mathrm{cos\alpha }=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \mathrm{and} \mathrm{sin\alpha }=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$
Then
$$\begin{gathered} \mathrm{y} =\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \left[\mathrm{cos\alpha } \mathrm{sin\omega t} + \mathrm{sin\alpha } \cos \mathrm{\omega t}\right] \\ \mathrm{y}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \sin \left(\mathrm{\omega t}+\mathrm{\alpha }\right) \end{gathered}$$

So, the motion is S.H.M. with amplitude $\sqrt{a^2+b^2}$

Hence the correct answer is option (d).
 

Answer:

Length of A and C are same.  Now A is given a transverse displacement. Since length of pendulums A and C is same time period is same and they will have same frequency of vibration. Due to which, a resonance will take place and the pendulum C will vibrate with maximum amplitude.
​Hence the correct answer is option (b).
 

Answer:

If $\omega$ is the angular velocity of the particle executing circular motion and $\theta$ is the angle as shown in the diagram, so
$\sin \theta =\frac{x}{OP}=\frac{x}{B} \left(\because OP=B\right)$

$$\begin{gathered} x=B\sin \theta \\ \mathrm{As} \theta =\omega t \\ x=B\sin \omega t \end{gathered}$$

$\Rightarrow x=B \sin \left(\frac{2\pi }{30}t\right) \left(\because \omega =\frac{2\pi }{T}=\frac{2\pi }{30}\right)$

Hence, the correct answer is option A.

Answer:

x varies between and +a and -a so the motion is oscillatory in nature.

We can also check for the periodicity of the motion by placing t+ T in place of time t  in the equation x = a cos (αt)² 

$\mathrm{x}(\mathrm{t}+\mathrm{T}) = \mathrm{acos}({\mathrm{\alpha }}^2{\mathrm{t}}^2 +{\mathrm{\alpha }}^2{\mathrm{T}}^2+2{\mathrm{\alpha }}^2\mathrm{tT}) \mathrm{is} \mathrm{not} \mathrm{equal} \mathrm{to} \mathrm{x}\left(\mathrm{t}\right)$

So it is not periodic.
Hence, the correct answer is option (c).


 

Answer:

Maximum velocity of the particle = aω = 30

Maximum acceleration of the particle = aω² = 60 

 $$\begin{gathered} \frac{\mathrm{Maximum} \mathrm{acceleration}}{\mathrm{Maximumm} \mathrm{velocity} } =\frac{{\mathrm{a\omega }}^2}{\mathrm{a\omega }}=\mathrm{\omega }= \frac{60}{30}= 2 \\ \mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}= 2 \\ \mathrm{T}= \mathrm{\pi } \mathrm{s} \end{gathered}$$
  
Hence, the correct answer is Option (a).

Answer:

In the above arrangement, the equivalent spring constant of the system is given as
k_eq = k₁ + k₂
The time period of the spring-block system is given as
$$\begin{gathered} T=2\pi \sqrt{\frac{m}{k_{eq}}}=2\pi \sqrt{\frac{m}{k_1+k_2}} \\ \nu =\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k_1+k_2}{m}} \end{gathered}$$

Considering the case when mass is connected to the two springs individually,
$$\begin{gathered} {\nu }_1=\frac{1}{2\pi }\sqrt{\frac{k_1}{m}} \\ {\nu }_2=\frac{1}{2\pi }\sqrt{\frac{k_2}{m}} \end{gathered}$$
Substituting value of k₁ and k₂ from above equations, 
$$\begin{gathered} \nu =\frac{1}{2\pi }\sqrt{4{\pi }^2{{\nu }_1}^2+4{\pi }^2{{\nu }_2}^2} \\ \nu =\sqrt{{{\nu }_1}^2+{{\nu }_2}^2} \end{gathered}$$

Hence, the correct answer is option (b).
 

Answer:

Earth takes a fixed period of time to complete a rotation about its axis i.e. it repeats it's motion after a fixed period of time, but it is not a to and fro motion.

∴ It's a periodic motion but not simple harmonic motion.

So, the correct options are option (a) and option (c). 
 

Answer:

The motion of a ball bearing inside a smooth curved ball is a to and fro motion as well as periodic as it will repeat itself in fixed time intervals. Here, the component of weight of the ball bearing parallel to incline of the bowl plays the role of restoring force. 

∴ It is a periodic as well as a simple harmonic motion. 

Hence the correct answers are option (a) and option (c). 
 

Answer:

Option a :  Displacements are in opposite direction at time 0 s and 2 s, hence phase is not same. 

Option b:  Displacements at time 2 s and 6 s are equal and in the same direction as well as the distance between them is of one exact time period, so they must be in the same phase.

Option c:  Though the displacements are the same at time 1 s and 7 s , but the particle is moving in opposite directions at these points, hence the phase is not the same.

Option d:  Again at time 1 s and 5 s are separated by one exact time period, so the phase must be the same.

Hence, the correct options are option (b) and option (d). 

Answer:

Standard equation of a particle undergoing S.H.M. is :  $x\mathit{=}a sin\mathit{(}\mathit{\omega }\mathit{t}\mathit{+}\mathit{\varphi }\mathit{)}$

As it's a sine function, it must be periodic.

Velocity is given by: $\frac{dx}{dt}=a\omega \cos \left(\omega t\mathit{+}\varphi \right)$

This gives velocity as a cosine function, so it must also be periodic.

Acceleration is given by: $\frac{dv}{dt}=-a{\omega }^2 \mathit{sin}\left(\omega t\mathit{+}\varphi \right)$

Acceleration is also a sine function, so it can not be constant.

$$\begin{gathered} \mathrm{Force}=\mathrm{mass}\times \mathrm{acceleration} \\ =m\mathit{\times }\mathit{\{}\mathit{-}{\mathit{\omega }}^{\mathit{2}}\mathit{a}\mathit{ }\mathit{sin}\left(\omega t+\varphi \right)\mathit{ }\mathit{\}} \\ =m\mathit{\times }\mathit{(}\mathit{-}{\mathit{\omega }}^{\mathit{2}}\mathit{x}\mathit{)} \\ =\mathit{-}m{\omega }^{\mathit{2}}x \end{gathered}$$

∴ Force acting is directly proportional to displacement from the mean position and opposite to it.

Hence, the correct options are option (a), option (b) and option (d).

Answer:

For a simple harmonic motion shown here:

Displacement from mean position at any time is given by 

⇒ $y\mathit{=}A\mathit{ }\mathit{cos}\omega t$ , where A is the maximum displacement from the mean position or the amplitude.

⇒$y\mathit{=}A\mathit{ }\mathit{cos}\left(\frac{2\pi }{T}t\right)$, where T is the time period of oscillation.

Velocity $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}\mathit{=}\mathit{-}A\omega \mathit{ }\mathit{sin}\left(\frac{2\pi }{T}t\right)$

Acceleration$={\omega }^{2}A \cos \left(\frac{2\pi }{T}t\right)$

Option a:
At time $t=\frac{3T}{4}$,  $\mathit{cos}\left(\frac{2\pi }{T}t\right)\mathit{=}\mathit{cos}\left(\frac{3\pi }{2}\right)\mathit{=}\mathit{0}\mathit{.}$
So, acceleration is zero, force is also 0. Option a is correct.

Option b:
At time $t\mathit{=}T$, acceleration$\mathit{=}{\omega }^{\mathit{2}}A$, which is the maximum value. Hence, this option is also correct.

Option c: 
At time $t\mathit{=}\frac{\mathit{T}}{\mathit{4}}$, velocity$\mathit{=}\omega A$ , which is the maximum value. Hence, this option is also correct.

Option d:
At time $t\mathit{=}\frac{\mathit{T}}{\mathit{2}}\mathit{ }\mathit{,}\mathit{ }\mathrm{displacement} y\mathit{ }\mathit{=}\mathit{-}A$, which corresponds to an extreme position. So, P.E. is maximum and K.E. is zero. Hence, this option is incorrect.

Hence, the correct answers are option (a), (b) and (c).


 

Answer:

We know that in S.H.M., total energy at any time is the sum of P.E. and K.E. 
Also at the extreme positions this sum is equal to maximum value of P.E. and at mean position this sum is equal to maximum value of K.E.

∴ Average total energy per cycle is equal to either Maximum K.E. or Maximum P.E. So, the first option is correct.

Average total energy is equally distributed among K.E. and P.E.

Hence, average kinetic energy per cycle can be said to be equal to half of its maximum kinetic energy. So, option b is also correct.

Mean velocity over a complete cycle is zero. So, option c is wrong.

Root mean square velocity$\mathit{=}\frac{{\mathit{v}}_{\mathit{m}\mathit{a}\mathit{x}}}{\sqrt{\mathit{2}}}$

So, option d is correct.

Correct options are a,b and d .

Answer:

Here, point O is mid-way between point A and point B. So, point O is the mean position.

Also recall that acceleration in S.H.M. is always towards the mean position.

Option a:
When the particle is 3 cm away from A going towards B, its velocity is positive, acceleration should be towards the mean position i.e. towards O, so it is also positive, hence force is also positive. This option is correct.

Option b:
When the particle is at point C and is going towards O, it is moving in positive direction as direction from point A to point B is given as positive direction in the question, so it's velocity is positive. This option is wrong.

Option c:
Here, the situation is opposite to option a, velocity is negative as the particle is moving towards A. Acceleration is towards mean position i.e. towards O but from left to right as O is not crossed from left, so acceleration is negative, hence the force. This option is correct.

Option d:
When the particle is at point B, acceleration should be towards O, i.e. negative, hence the force is also negative. So, this option is also correct.

Correct options: a,c,d

Answer:

Velocity is zero at extreme positions and speed is maximum at mean positions.

So, points with zero velocity: A,C,E,G

Points with maximum speed: B,D,F,H
 

Answer:

According to the question both the springs are identical and k = K.
When the mass is displaced from equilibrium position by a distance x towards right:

1. The right spring gets compressed by x developing a restoring force of Kx towards left of the block.

2. The left spring gets stretched by x developing a restoring force of Kx  again towards the left of the block.

∴ Total restoring force is 2Kx  towards the left of the block.
 

Answer:

The two basic characteristics of a simple harmonic motion are:

1. The magnitude of acceleration is proportional to the magnitude of displacement.

2. The direction of acceleration is opposite to displacement from mean position. 

Answer:

Motion of a simple pendulum is said to be simple harmonic when the displacement of the bob of the pendulum is very small from the mean position, such that sinθ ≅ θ.

Answer:

For a particle executing simple harmonic motion,
Displacement, $x=a\sin \left(\omega t+\varphi \right)$
Velocity, $v=\frac{dx}{dt}=a\omega \cos \left(\omega t+\varphi \right)$
Thus, Maximum velocity, ${\left|v\right|}_{max}=a\omega$
Acceleration, $a=\frac{d^{2}x}{d{t}^2}=-a{\omega }^2\sin \left(\omega t+\varphi \right)$
Thus, Maximum acceleration, ${\left|a\right|}_{max}=a{\omega }^2$
Now, the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator is given as
$\frac{{\left|a\right|}_{max}}{{\left|v\right|}_{max}}=\frac{a{\omega }^2}{a\omega }=\omega$
 

Answer:

Distance travelled by the oscillator in one time period = 4 times amplitude

$\frac{\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{the} \mathrm{oscillator} \mathrm{in} \mathrm{one} \mathrm{time} \mathrm{period}}{\mathrm{Amplitude}} = 4$

Answer:

The particle  is moving in anti-clockwise direction as shown in the diagram. Thus, the projection will move from right towards left and hence the sign of the velocity of the point P′ is negative.

Answer:

For a particle executing S.H.M.:

Assuming displacement is given by:   $x\mathit{=}A\mathit{ }\mathit{cos}\omega t$

∴ Phase of displacement $\mathit{=}\omega t$

Velocity:  $\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\mathit{=}\mathit{-}A\omega \mathit{ }\mathit{sin}\omega t\mathit{ }\mathit{=}\mathit{ }\mathit{-}A\omega \mathit{ }\mathit{cos}(\omega t+\frac{\pi }{2})$

∴ Phase of velocity $\mathit{=}\omega t\mathit{+}\frac{\mathit{\pi }}{\mathit{2}}$

∴ Phase difference between velocity and displacement $\mathit{=}\frac{\mathit{\pi }}{\mathit{2}}$

Answer:

Following is graph showing variation of potential energy, kinetic energy and total energy of a simple harmonic oscillator with displacement:


Here,
 $$\begin{gathered} \mathrm{U}\left(\mathrm{x}\right)=\mathrm{Potential} \mathrm{energy} \\ \mathrm{K}\left(\mathrm{x}\right)=\mathrm{Kinetic} \mathrm{energy} \\ \mathrm{E}=\mathrm{U}\left(\mathrm{x}\right)+\mathrm{K}\left(\mathrm{x}\right)=\mathrm{Total} \mathrm{energy} \end{gathered}$$

Answer:

Time period of simple pendulum, $T=2\pi \sqrt{\frac{l}{g}}$
Second's pendulum is a simple pendulum with time period 2 s.
If l_e and l_m is the length of a second’s pendulum on the surface of Earth and moon respectively, then 
$\frac{l_e}{l_m}=\frac{g_e}{g_m}=6$
Where, acceleration due to gravity on earth(g_e) = 6 times the acceleration due to gravity on moon(g_m)
$\Rightarrow l_m=\frac{1}{6} \mathrm{m}$

Answer:

In equilibrium,
Mg = T + T
Mg = 2T
If hanging mass moves by distance l, then mass spring elongates by 2l, i.e.
T = F
Mg = 2(2kl) = 2k(2l)                                            (1)
When mass is displaced through distance y downwards, the restoring force on mass M is given as
F = Mg − 2k(2l+2y)
F = Mg − 2k(2l )− 4ky
Using (1),
F = Mg − Mg − 4ky
$$\begin{gathered} M\frac{d^{2}y}{d{t}^2}=-4ky \\ \frac{d^{2}y}{d{t}^2}=-\frac{4k}{M}y \end{gathered}$$
Comparing it with standard equation: $\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}y$
$$\begin{gathered} \omega =\sqrt{\frac{4k}{M}} \\ \Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{M}{4k}} \end{gathered}$$
 

Answer:

$$\begin{gathered} y = \sin \omega t – \cos \omega t \\ y =\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \omega t – \frac{1}{\sqrt{2}}\cos \omega t\right) \\ y =\sqrt{2}\left( \sin \omega t \cos \frac{\pi }{4} – \cos \omega t \sin \frac{\pi }{4}\right) \\ y =\sqrt{2} \sin \left(\omega t-\frac{\pi }{4}\right) \end{gathered}$$

Comparing with standard equation: $y=a\sin \left(\omega t+\varphi \right)$, we have
$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{T} \\ T=\frac{2\mathrm{\pi }}{\omega } \end{gathered}$$

Answer:

According to the question,
$\mathrm{P}.\mathrm{E}.=\frac{1}{2}\left(\mathrm{T}.\mathrm{E}.\right)$
Where T.E. is the total energy or maximum energy of the oscillator

If x is the required displacement and A is the amplitude of the oscillation, then we have
$$\begin{gathered} \frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}\left(\frac{1}{2}m{\omega }^2{A}^2\right) \\ {x}^2=\frac{A^2}{2} \\ x=\pm \frac{A}{\sqrt{2}} \end{gathered}$$

Answer:

U(x) = U₀ (1 – cos αx)
$$\begin{gathered} F=-\frac{dU}{dx}=-\frac{d}{dx}\left[U_0 –U_0\cos \alpha x\right] \\ =-\alpha U_0\sin \alpha x \end{gathered}$$

For small $\alpha x, \sin \alpha x\approx \alpha x$
$F=-U_0 {\alpha }^2 x$

Comparing it with $F=-kx$

$k=U_0 {\alpha }^2$
Now, Time period is given as
$$\begin{gathered} T=2\pi \sqrt{\frac{m}{k}} \\ =2\pi \sqrt{\frac{m}{U_{0 }{\alpha }^2}} \end{gathered}$$

Answer:

Given:
Mass, m = 2 kg
Spring constant, k = 50 Nm^–1
Amplitude, A = 5 cm = 0.05 m

According to the relation,
$$\begin{gathered} \omega =\sqrt{\frac{k}{m}} \\ \omega =\sqrt{\frac{50}{5}}=5 \mathrm{rad}/\mathrm{s} \end{gathered}$$
Thus, displacement is given as
$$\begin{gathered} \mathrm{x}=\mathrm{Asin\omega t} \\ \mathrm{x}=\left(0.05\sin 5\mathrm{t}\right) \mathrm{m} \\ \mathrm{Or} \mathrm{x}=\left(5\sin 5\mathrm{t}\right) \mathrm{cm} \end{gathered}$$

Answer:

Angular displacement for both the pendulums is given as
$$\begin{gathered} {\theta }_1={\theta }_0\sin \left(\omega t+{\delta }_1\right) \\ {\theta }_2={\theta }_0\sin \left(\omega t+{\delta }_2\right) \end{gathered}$$
For first pendulum,
$$\begin{gathered} 2°=2°\sin \left(\omega t+{\delta }_1\right) \\ \sin \left(\omega t+{\delta }_1\right)=1 \\ \omega t+{\delta }_1=90° ... \left(1\right) \end{gathered}$$

For second pendulum,
$$\begin{gathered} -1°=2°\sin \left(\omega t+{\delta }_2\right) \\ \sin \left(\omega t+{\delta }_2\right)=-\frac{1}{2} \\ \omega t+{\delta }_2=-30° ...\left(2\right) \end{gathered}$$

From (1) and (2)
${\delta }_1-{\delta }_2=120°$
 

Answer:

Mass of the person, m = 50 kg
Frequency, f  = 2 s^-1
Amplitude, A = 5 cm = 0.05 m
Angular velocity, $\omega =2\pi f=2\pi \times 2=4\pi s^{-1}$
(a) Yes, there will be change in weight of the body as measured by the weighing machine during the oscillation. Weight of the person will change from maximum to minimum value as calculated below:

(b) When the person is at topmost position during the oscillation:
Weight (W) shown by the weighing machine= normal reaction (N)
$$\begin{gathered} mg-N=ma \\ \Rightarrow N=mg-ma=mg-mA{\omega }^2 \left(\mathrm{acceleration}, a=A{\omega }^2\right) \\ \Rightarrow N=50\times 9.8-50\times 0.05\times {\left(4\times \pi \right)}^2 \\ =95.6 \mathrm{N} (\mathrm{Minimum} \mathrm{value}) \end{gathered}$$

When the person is at lowest position during the oscillation:
$$\begin{gathered} N-mg=ma \\ \Rightarrow N=mg+ma=mg+mA{\omega }^2 \left(\mathrm{acceleration}, a=A{\omega }^2\right) \\ \Rightarrow N=50\times 9.8+50\times 0.05\times {\left(4\times \pi \right)}^2 \\ =884.3 \mathrm{N} (\mathrm{Maximum} \mathrm{value}) \end{gathered}$$


                                  


 

Answer:

(a) At its lowest position:
The body displaces to 4 cm from the initial position. The body will oscillate between the lowest and the initial position.
The mean position is in the mid of  its path of oscillation.
So, the amplitude = distance of lowest/topmost position from its mean position of the SHM
                             =$\frac{4}{2}=2 \mathrm{cm}$

(b) Frequency of the oscillation,
$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{m}} \left(m=\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{body}, k=\mathrm{spring} \mathrm{constant}\right)$
At mean position,
$$\begin{gathered} mg=kx \left(x=\mathrm{elongation} \mathrm{in} \mathrm{the} \mathrm{spring}=2 \mathrm{cm}=0.02 \mathrm{m}\right) \\ \Rightarrow k=\frac{mg}{0.02} \end{gathered}$$
$$\begin{gathered} \mathrm{So}, \mathrm{frequency} \\ f=\frac{1}{2\pi }\sqrt{\frac{{\displaystyle \frac{mg}{0.02}}}{m}}= \frac{1}{2\times 3.14}\times \sqrt{\frac{9.8}{0.02}}=3.5 {\mathrm{s}}^{-1} \end{gathered}$$
 

Answer:

The log of wood is pressed and released.
Let the vertical displacement of the wood at the equilibrium position be x_o .
At equilibrium, the buoyant force acting on the wood must be equal to its weight,
$mg=V\rho g=A{x}_o\rho g \left(\mathrm{Volume},V=A{x}_o\right)$
When it is further pressed to x from equilibrium position , then the buoyant force:
$V\text{'}\rho g=A\left(x+x_o\right)\rho g \left[\mathrm{Volume},V\text{'}=A\left(x+x_o\right)\right]$
Hence, the net restoring force = buoyant force – weight
 $$\begin{gathered} \Rightarrow R=\left(A\rho g\right)x=kx \left(where, k=A\rho g\right) \\ \Rightarrow R\propto x \\ \mathrm{So}, \mathrm{the} \mathrm{time} \mathrm{period} \mathrm{of} \mathrm{the} \mathrm{oscillation}: \\ T\mathit{=}\mathit{2}\pi \sqrt{\frac{\mathit{m}}{\mathit{k}}}\mathit{=}\mathit{2}\pi \sqrt{\frac{\mathit{m}}{\mathit{A}\mathit{\rho }\mathit{g}}} \end{gathered}$$


 

Answer:

Let l be the length of one arm of the tube in either side as shown in the figure.
Taking a small length of the tube dx at a height x from the base as shown in the figure.

Mass of the liquid in this small tube length, dm = Adxρ       (ρ = density of the liquid, A=area of cross-section of the tube)
Potential energy of this small mass = Aρgxdx
Now potential energy of the liquid in the left arm of the tube:
$$\begin{gathered} P{E}_{left}={\int }_0^{h_1}A\rho gxdx=A\rho g\frac{{h_1}^2}{2} \\ =A\rho g\frac{{\left(l\sin {45}^o\right)}^2}{2}=\frac{A\rho g{l}^2}{4} \left(\mathrm{In} \mathrm{the} \mathrm{given} \mathrm{diagram}, h_{\mathit{1}}=l\sin {45}^{\mathrm{o}}\right) \\ \mathrm{Similarly}, \\ P{E}_{left}={\int }_0^{h_2}A\rho gxdx=A\rho g\frac{{h_2}^2}{2}=\frac{A\rho g{l}^2}{4} \end{gathered}$$
So total potential energy of the system,
$TPE=P{E}_{left}+P{E}_{right}=\frac{A\rho g{l}^2}{2}$

When small pressure difference is created between the liquids in the two arms:
Let the change in liquid level along the tube in left side be y.
So, the length of the liquid in left arm is l–y and in the right arm is l + y.
So, the total potential energy of the system in this situation:
$TPE\text{'}=P{E}_{left}+P{E}_{right}=\frac{A\rho g}{2}\left[{\left(l-y\right)}^2+{\left(l+y\right)}^2\right]$
Change in total potential energy in these two situations:
$$\begin{gathered} TPE\text{'}-TPE=P{E}_{left}+P{E}_{right}=\frac{A\rho g}{2}\left[{\left(l-y\right)}^2+{\left(l+y\right)}^2-l^2\right]=\frac{1}{2}A\rho g\left(2y^2+l^2\right) \\ \mathrm{Change} \mathrm{in} \mathrm{kinetic} \mathrm{energy} =\frac{1}{2}\times m{v}^2=\frac{1}{2}\times A\rho \mathit{\left(}\mathit{2}l\mathit{\right)}{v}^2\mathit{ }\mathit{=}A\rho l{v}^2\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{(}\mathit{total}\mathit{ }\mathit{length}\mathit{ }\mathit{of}\mathit{ }\mathit{the}\mathit{ }\mathit{liquid}\mathit{ }\mathit{column}\mathit{=}\mathit{2}l\mathit{)} \\ \mathrm{So}, \mathrm{by} \mathrm{applying} \mathrm{conservation} \mathrm{of} \mathrm{energy}: \\ \mathrm{Change} \mathrm{in} \mathrm{total} \mathrm{energy}=0 \\ \frac{1}{2}A\rho g\left(2y^2+l^2\right)+A\rho l{v}^2=0 \\ \mathrm{Now}, \mathrm{diffentiating} \mathrm{both} \mathrm{side} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{time} \mathrm{we} \mathrm{get}: \\ \frac{d}{dt}\mathit{[}\frac{\mathit{1}}{\mathit{2}}A\rho g\mathit{(}\mathit{2}{y}^{\mathit{2}}\mathit{+}{l}^{\mathit{2}}\mathit{)}+A\rho l{v}^{\mathit{2}}\mathit{]}=0 \\ \Rightarrow \mathit{\left[}A\rho g\mathit{(}\mathit{2}y\frac{dy}{dt}\mathit{+}0\mathit{)}\mathrm{+}2A\rho lv\frac{dv}{dt}\mathit{\right]}\mathit{=}\mathit{0} \\ \mathit{\Rightarrow }A\rho g\mathit{\left(}\mathit{2}\mathit{y}\mathit{v}\mathit{\right)}+2A\rho lv\frac{dv}{dt}=0 \left(\mathrm{as}, \mathrm{velocity}=\frac{dy}{dt}=v \mathrm{and} \mathrm{acceleration}=\frac{dv}{dt}=a\right) \\ \Rightarrow gy+la=0 \\ \Rightarrow a=-\frac{g}{l}y \\ \mathrm{Compairing} \mathrm{this} \mathrm{equation} \mathrm{with} \mathrm{standard} \mathrm{equation} \mathrm{of} \mathrm{SHM}, \\ \Rightarrow \mathrm{a}=-{\omega }^{2}y\mathit{ }\mathit{ } \\ \mathrm{So}, \omega = \sqrt{\frac{g}{l}} \end{gathered}$$
 $\mathrm{Hence}, \mathrm{time} \mathrm{period}, T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{g}}$





 

Answer:

Let a tunnel is dug through the centre of the Earth.
Let mass of the earth be M and its radius be R.
Suppose at an instant t the particle in the tunnel is at a distance x from the centre of the earth.
Acceleration due to gravity acting on the body of mass m at that depth:
$g\text{'}=g\left(1-\frac{R-x}{R}\right)=\frac{gx}{R}$
Force acting on the body of mass m at that depth,
$F=m\frac{gx}{R}=kx \mathrm{towards} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{earth}$
$F\propto x, Where, k =\frac{mg}{R}=\frac{GmM}{R^3} \left(\mathrm{As}, g=\frac{GM}{R^2}\right)$
This force always pulls the body towards the centre of the earth.
Hence, it execute simple harmonic motion with time period,
​$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{{\displaystyle \frac{mg}{R}}}}=2\pi \sqrt{\frac{R}{{\displaystyle g}}}$

 

Answer:

Given: time period, T = 1 s 
Length = l
Amplitude of the pendulum = ${\theta }_o$
So, angular frequency $\omega =\frac{2\pi }{T}=\frac{2\pi }{1}=2\pi {\mathrm{s}}^{-1}$
It is also given that  we should assume θ₀ to be small so that sinθ₀ =θ₀ and cosθ₀ =1.

Now the angular displacement of the bob of the pendulum at any instant t is given by:
$\theta ={\theta }_o\sin \omega t={\theta }_o\sin 2\pi t$    ...(1)




Let the angle made by the pendulum with the vertical at time t = 0 be $\theta ={\theta }_o$.
Now at time, t₁  l et  $\theta =\frac{{\theta }_o}{2}$
Hence,
 $$\begin{gathered} \frac{{\theta }_o}{2}={\theta }_o\sin 2\pi t_1 \\ \Rightarrow \sin 2\pi t_1=\frac{1}{2} \\ \Rightarrow 2\pi t_1=\frac{\pi }{6} \\ \Rightarrow t_1=\frac{1}{12} \mathrm{s} \end{gathered}$$
Differentiating equation (1) both side with respect to time t we get:
$$\begin{gathered} \mathrm{Angular} \mathrm{velocity}=\frac{d\theta }{dt}=2\pi {\theta }_o\cos 2\pi t ...\left(2\right) \\ \mathrm{Now}, \mathrm{at} t =\frac{1}{12} \mathrm{s} \\ \mathrm{Angular} \mathrm{velocity}=\frac{d\theta }{dt}\mathit{=}2\pi {\theta }_o\cos \mathit{(}\mathit{2}\mathit{\pi }\mathit{\times }\frac{\mathit{1}}{\mathit{12}}\mathit{)}=\sqrt{3}\pi {\theta }_o \\ \mathrm{Velocity} \mathrm{of} \mathrm{the} \mathrm{bob}, v=\mathrm{Angular} \mathrm{velocity}\times l=\sqrt{3}\pi {\theta }_{o}l \end{gathered}$$
Now horizontal and vertical component of the bob is:
​$$\begin{gathered} v_x=v\cos \frac{{\theta }_o}{2}=\sqrt{3}\pi {\theta }_{o}l\cos \frac{{\theta }_o}{2} \\ {v}_y=v\sin \frac{{\theta }_o}{2}=\sqrt{3}\pi {\theta }_{o}l\sin \frac{{\theta }_o}{2} \\ \mathrm{Now} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{bow} \mathrm{when} \mathrm{bob} \mathrm{snaps}, \\ H\mathit{\text{'}}=H+l\mathit{-}l\cos \frac{{\mathit{\theta }}_{\mathit{o}}}{\mathit{2}}\mathit{=}H\mathit{+}l\left(1-\cos \frac{{\theta }_o}{\mathit{2}}\right) \\ \mathrm{Let} \mathrm{it} \mathrm{takes} \mathrm{time} t \mathrm{to} \mathrm{reach} \mathrm{the} \mathrm{ground}: \\ H\mathit{\text{'}}=v_{\mathit{y}}t+\frac{1}{2}g{t}^{\mathit{2}} \\ \mathrm{Putting} \mathrm{all} \mathrm{the} \mathrm{values} \mathrm{and} \mathrm{taking} \mathrm{approximation} \mathrm{as} {\theta }_{\mathrm{o}} \mathrm{is} \mathrm{very} \mathrm{very} \mathrm{small}, \\ \mathrm{solving} \mathrm{for} t \mathrm{we} \mathrm{get}: \\ t=\sqrt{\frac{\mathit{2}\mathit{H}}{\mathit{g}}} \\ \mathrm{Now} \mathrm{horizontal} \mathrm{range} \mathrm{of} \mathrm{the} \mathrm{bob}: \\ R=v_{\mathit{x}}t\mathit{=}\sqrt{3}\pi {\theta }_{\mathit{o}}l\cos \frac{{\mathit{\theta }}_{\mathit{o}}}{\mathit{2}}\mathit{\times }\sqrt{\frac{\mathit{2}\mathit{H}}{\mathit{g}}}=\sqrt{\frac{6H}{g}}\pi {\theta }_{o}l \left(\mathrm{Using} \mathrm{the} \mathrm{approximation} {\theta }_{\mathrm{o}} \mathrm{is} \mathrm{very} \mathrm{very} \mathrm{small}\right) \\ \mathrm{So} \mathrm{distance}, \mathrm{from} \mathrm{point} \mathrm{A}, \\ \mathrm{x}=l\sin \frac{{\mathit{\theta }}_{\mathit{o}}}{\mathit{2}}\mathit{-}\sqrt{\frac{\mathit{6}\mathit{H}}{\mathit{g}}}\pi {\theta }_{\mathit{o}}l\mathit{=}l\times \frac{{\mathit{\theta }}_{\mathit{o}}}{\mathit{2}}\mathit{-}\sqrt{\frac{\mathit{6}\mathit{H}}{\mathit{g}}}\pi {\theta }_{\mathit{o}}l={\theta }_{\mathit{o}}l\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{-}\pi \sqrt{\frac{\mathrm{6}H}{g}}\right) \end{gathered}$$