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When a source is sounded, it generally vibrates in more than one mode and therefore, emits tones of different frequencies. The tone of lowest frequency is called the fundamental note and the tones of higher frequencies are called overtones.

Necessary properties of the medium for wave propagation : 

• Elasticity : So that particles can return to their mean position, after having been disturbed.
• Inertia : So that particles can store energy and overshoot their mean position.
• Minimum friction amongst the particles of the medium. 
• Uniform density of the medium.

The frequency will not change. Because frequency depends on length of air column above the liquid surface in the tube.

Properties of elasticity and inertia.

Type of wave motion is determined by

(i) nature of the medium,

(ii) mode of excitation of wave motion.

Electromagnetic waves, e.g., light waves need no medium for propagation.

B) 126 J

Explanations:

\(Q_p=\mu C_p\Delta T \,and\, Q_v = \mu C_v\Delta T\)

\(\Rightarrow \frac{QV}{Qp}=\frac{CV}{CP}=\frac{\frac{3}{2}R}{\frac{5}{2}R}=\frac{3}{5}\)

\([\therefore(CV)_{mono}=\frac{3}{2}R,(C_p)_{mono}=\frac{5}{2}R]\)

\(\Rightarrow QV=\frac{3}{5}\)\(\times(\Delta Q)P\)

\(=\frac{3}{5}\times210 = 126J\)

Data : A = 5 cm, T = 6 s, x = 2.5 cm 

Since the particle starts from the mean position, its epoch, α = 0.

∴ The equation of motion is x = A sin ωt

∴ The required phase of the particle,

ω = sin^-1 \(\frac xA\)

= sin^-1 \(\frac{2.5}5\) = sin^-1 \(\frac 12\) = \(\frac {\pi}6\) rad

(b) and 

(c) are SHMs 

and (d) are periodic, but not

SHMs

During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis. 

An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time. 

The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic. 

1. Periodic but not SHM (Simple Harmonic Motion) 

2. SHM 

3. SHM 

4. Periodic but not SHM

Here, the maximum displacement = Amplitude (A) = 2 cm

(a) If the time t = 0 at x = 0, the displacement can be expressed as the function of t.

i.e., x(t) = A sin ωt

But, ω = √{k/m} = √{1200/3} = 20 rad s^-1

i.e., x(t) = 2 sin 20 t

(b) If the time t = 0 at x = 2 cm (i.e., positive extreme position), the displacement is expressed as the cosine function.

i.e., x(t) = 2 cos 20 t

(c) If the time t = 0 at x = -2 cm (i.e., negative extreme position), the displacement is expressed as the cosine function given by

x(t) = -cos 20 t

The above functions have same values of frequency and amplitude but they differ in phase.

Only (a) and (c) represent periodic motion

(b) and (c) represent S.H.M. in particular

(a) represents the periodic motion

(b) represents a superposition of a number of S.H.M. motions and is a periodic motion not necessarily S.H.M.

U(x) = Uo(1 − cos αx)

Differentiating both sides with respect to x

\(\frac{dU(x)}{dx}\)= U_o[0 + α sin αx ] = U_oα sin αx

∴ F = − \(\frac{dU(x)}{dx}\) = −U_oα sin α x 

When oscillations are small, sin θ ≈ θ

or sin αx = αx 

∴ F = −U_oα(αx) = −U_oα²x

∴ F = −(U_oα²)  (i)

We know that F = −kx  (ii)

k = U_oα²  {From (i) & (ii)}

∴ T = \(2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{m}{U_o\alpha^2}}\)

The sources of restoring force is as follows:

(i) Simple pendulum : gravity.

(ii) Spring : elasticity

(iii) Column of mercury in U tube : weight

The frequency of external periodic force is different from the natural frequency of the oscillator in case of forced oscillation but in resonance two frequencies are equal. 

Let the full spring be the combination of two springs in series, each of force constant k. then in case (I), the effective spring constant (k₁) is given by

k₁ = \(\frac{k\times k}{k\times k}=\frac{k}{2}\) 

Frequency of oscillation,

f₁ = \(\frac{1}{2\pi}\sqrt{\frac{k_1}{m}}\)

= \(\frac{1}{2\pi}\sqrt{\frac {k}{2m}}\)

In case (ii); when the spring is cut to one-half, the effective spring constant k₂ = k. Frequency of oscillation,

f₂ = \(\frac{1}{2\pi}\sqrt{\frac{k_2}{m}}\)

= \(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\) 

∴ \(\frac{f_2}{f_1}=\sqrt 2\)

The frequency of oscillations of mass m and force constant k is, f1​ = 1/2π under root k/​m​​
when the spring is cut to one-half of its length, its force constant is doubles (2k).
Then frequency of oscillation of mass m will be 

f2​=1/2π under root ​2k/m​​ =​ f2/f1 ​​= under root2​`

As the pendulum oscillates, it drags air along it. Therefore, its K.E. is dissipated in overcoming viscous drag due to air and hence its amplitude goes on decreasing.

1. Periodic time does not change. T is independent of amplitude.

2. Total energy, T.E = 1/2 mω² A²

A is doubled ⇒ T.E_new = 4 TE_old

3. Maximum velocity, V_max = ωA 

A is doubled ⇒ V_max is doubled.

A) Reduce the mass to one fourth

Explanations:

The frequency of spring oscillator is

\(v=\frac{1}{2\pi}\sqrt\frac{k}{m}\)

\(\implies v\propto \sqrt\frac{1}{m}\)

Hence \(\frac {v_1}{v_2}=\sqrt\frac{m_2}{m_1}\)

or \(\frac{m_2}{m_1}=(\frac{v_1}{v_2})^2=(\frac{v}{2v})^2=\frac{1}{4}\)

\(m_2=\frac{m_1}{4}\)