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In case of a simple pendulum, k is directly proportional to m. Hence ratio of m / k is a constant. Hence time period doesn’t depend on mass.

In a solid:

Speed of transverse wave, \(v=\sqrt{\frac{η}{ρ}}\)

Where η is modulus of rigidity and ρ is density of material of solid.

In a stretched string: Here \(v=\sqrt{\frac{T}{m}}\)

Where T is tension in the string and m is mass per unit length of the string.

E = \(\frac{1}{2}ka^2\), where a is the amplitude and k- force constant.

Dimensional formula = [ML²T ^-2 ].

Free oscillation 

(i) The oscillation of a particle with fundamental frequency under the influence of restoring force are defined as free oscillations 

(ii) The amplitude, frequency and energy of oscillation remains constant 

(iii) Frequency of free oscillation is called natural frequency. 

Damped oscillation : 

(i) The oscillation of a body whose amplitude goes on decreasing with time are defined as damped oscillation. 

(ii) Amplitude of oscillation decreases exponentially due to damping forces like frictional force, viscous force, hystersis etc. 

Here the trolley is displaced slightly through and the other spring is compressed. Therefore. a net restoring force F comes into play such that

F = -k_1y - k₂y

= -(k₁ + k₂)y

⇒ ma = -(k₁ + k₂)y

or, a = -({k₁ + k₁}/{m}) x y   ...(i)

Comparing equation (i) with the equation,

a = -w²y,

We get, ω = √{k₁ + k₁/m}

Therefore, T = 2π/ω

⇒ T = 2π√{m/k₁ + k₂}

Here, k₁ = k₂ = 600 Nm^-1, m = 3 kg

(a) T = 2 x 3.14 x √{3/600 + 600} = 0.0314 s

(b) When y = 5 cm = 5 x 10^-2 m

The potential energy stored,

U = 1/2 (k₁ + k₂)y₂ = 1/2 (600 + 600) x (5 x 10^-2)²

= 1.5 J

Now, maximum velocity is attained, when whole of the potential energy is converted into kinetic energy.

i.e., 1/2 mv²_max = 1.5

Here, v_max = √{(1.5 x 2)/(m)}

= √{(1.5 x 2)/(3)} = 1 ms^-1

(c) If the trolley comes to rest due to damping forces, the kinetic energy is converted into heat energy.

Thus the heat energy produced = 1.5 J.

y = a sin wt + b cos wt

a = r cosθ and b = r sinθ

y = r sin(wt = θ)   ....(i)

a² + b² = r²(sin²θ + cos²θ) = r²

or, r = √{a² + b²}   (taking positive sign)

Putting this value in (i), we have

y = √{(a² + b²) sin (wt = θ)}

Therefore the given equation represents a S.H.M.

Amplitude of motion, r = √{a² + b²}

Length of the path, 2r = 2√{(a² + b²)}

Various physical quantities in S.H.M. at different position :

A) 3T

Explanations:

The periodic time of a simple pendulum is given by

\(T=2\pi\sqrt{\frac{l}{g}}\)

\(T_1 =T,l_1=l,l_1=9l\)

\(\frac{T_1}{T_2}=\sqrt\frac{l_1}{l_2}=\sqrt\frac{1}{9}=\frac{1}{3}\)

\(T_2=3T_1=3T\)

Speed of sound in a gas is \(v=\sqrt{\frac{\gamma P}{ρ}}\) …(i)

According to kinetic theory of gases, root mean square velocity (c) of molecules of gas is obtained from the relation

\(P=\frac{1}{3}ρc^2,\)  \(c = \sqrt{\frac {3P}{ρ}}\) …(ii)

Dividing (i) by (ii), 

\(\frac{v}{c}=\sqrt{\frac{\gamma}{3}}\) 

Or  \(v=c \sqrt{\frac{\gamma}{3}}\) 

This is the required solution.

As we move up, the pressure (P) of air and density of air (ρ), both decreases.

As v = \(\sqrt{\frac{\gamma Rt}{M}}\), therefore, velocity of sound will not change.

The person would be hearing nothing if he is moving away from the source with the speed of sound. It is so, because the relative velocity of sound waves with respect to person is zero. Therefore, the sound waves cannot strike the drum of the person’s ears and hence, no sensation of hearing is produced.

As speed of sound in water is roughly four times the speed of sound in air, therefore

μ = \(\frac {sin \,i}{sin\, r}=\frac{v_a}{v_w}\)

= \(\frac{1}{4}\) = 0.25

For refraction,

r_max = 90º

∴ (sin i)_max =0.25

∴ i_max ≈ 14º . That is why most of sound produced in air and falling at ∠i>14º gets reflected in air and person deep inside water cannot hear the sound.

(i) v ∝ \(\sqrt T\), i.e., speed of sound varies directly as the square root of the temperature in kelvin.

(ii) v ∝ \(\frac{1}{\sqrt p}\) , i.e., speed of sound in a gas varies inversely as the square root of density of gas.

(iii) Sound travels faster in moist air than in the dry air.

(iv) Change in pressure has no effect on speed of sound.

(v) Wind has no effect on speed of θ is 90º. Sound travels at higher speed if θ is less than 90º and at lower speed if θ is more than 90º.

The P.E. of a particle executing S.H.M. is given by:

\(E_P=\frac{1}{2}m\omega^2y^2.\)

E_P is maximum when y = r = amplitude of vibration, i.e., the particle is passing from the extreme position and is minimum when y = 0, i.e., the particle is passing from the mean position.

The K.E. of a particle executing S.H.M. is given by;

\(E_k=\frac{1}{2}m\omega^2(r^2\,-\,y^2)\)

E_K is maximum when y = 0, i.e., the particle is passing from the mean position and E_K is minimum when y = r, i.e., the particle is passing from the extreme position.

When the displacement of a particle executing S.H.M. is y, then its

K.E. = \(\frac{1}{2}\)mω²(A² − y²)

And P.E. = \(\frac{1}{2}m\omega^2y^2\)

If K.E. = P.E.

then, \(\frac{1}{2}m\omega^2(A^2\,-\,y^2)=\frac{1}{2}m\omega^2y^2\)

Or 2y² = A²

Or y = \(\frac{A}{\sqrt2}\)

PE of oscillator, U = \(\frac{1}{2}mw^2y^2\)  

{y = displacement Maximum – energy of oscillator, E = \(\frac{1}{2}mw^2A^2\)}

U = \(\frac{1}{2}E\)

or \(\frac{1}{2}mw^2y^2\) = \(\frac{1}{2}mw^2A^2\)

or y² = \(\frac{A^2}{2}\)

or y = \(\pm\frac{A}{\sqrt 2}\)

Here, v₀ =  ωA, a₀ = ω²A = (ωA)ω = v₀ω

or v₀ω = a₀

ω = \(\frac{a_0}{v_0}\)

Thus  A = \(\frac{v_0}{ω}=\frac{v_0}{\frac{a_0}{v_0}}=\frac{v^2_0}{a_0}\)

\(\omega\) = \(\sqrt{\frac{k}{m}}\) = \(\sqrt{\frac{4.84}{0.98}}\) = 2.22 rad / sec

(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive.

(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative.

(d) The sign of acceleration and force on the particle when it is at point B is negative.

Correct option is: (A) \(\cfrac{\sqrt{3}}{2}\)A

Given that: v² = α − βy² …(i)

Differentiating it with respect to time t, we have

2v\(\frac{dv}{dt}\) = −2βy\(\frac{dy}{dt}\)

or  \(\frac{dv}{dt}\) = −βy…(ii)

It means acceleration

a =\(\frac{dv}{dt}\)= −βy

As a and y have –ve sign shows that acceleration is directed towards mean position, so, if the particle is left free, it will execute S.H.M.

Here ω² = β

ω = \(\sqrt{\beta}\) 

∴ Time period, T = \(\frac{2\pi}{\omega}\) = \(\frac{2\pi}{\sqrt \beta}\) 

We know that, v = 0, when y = r from (ii)

0 = α – βr²

or r = \(\sqrt{\frac{\alpha}{\beta}}\) 

i.e., amplitude, r = \(\sqrt{\frac{\alpha}{\beta}}\)

A particle is said to have SHM if the acceleration of the particle is directly proportional to its displacement from the mean position and directed towards the mean position. 

E.g: 

1. Vertical oscillations of a loaded spring. 

2. Oscillations of bob of a simple pendulum. 

3. Vibrations of string of musical instruments. 

4. Motion of air particles during the propagation of sound waves. 

5. Vibration of a tuning fork.

Velocity of a particle executing SHM is v = \(ω\sqrt {OA^2+AC^2}\)

Acceleration of a particle executing SHM is a = – ω²y y is the displacement of a particle from its mean position in t seconds. A is the amplitude and ω is the angular frequency.

On the surface of the earth,

g = 9.38 ms^-2, T = 3.5 s.

Using T = 2π√{l/g}

⇒ l = T²g/4π²

or, l = {(3.5)² x 9.8}/{4 x 9.87} = 3.04 m

On the surface of the moon,

g = 1.7 ms^-2, l = 3.04 m

Hence T = 2π√{l/g}

= 2 x 3.14 x √{3.04/1.7} = 8.4 sec.