Explore topic-wise InterviewSolutions in .

Projection of P on x-axis

\(x=A\,cos(wt\,+\,\phi)\)

\(v_x=\frac{dx}{dt}=-Aw\, sin(wt + \phi)\)

∴ sign of velocity is negative

0º if both are moving in same direction and 180º or π radians if moving in opposite directions.

When the sign in the force equation is changed from ‘− ve’ to ‘+ ve’, the force and hence acceleration will not be opposite to displacement. Due to which the particle will not oscillate but will be accelerated in the direction of displacement. As a result of it, the motion will become a linearly accelerated motion.

We know that time period of a simple pendulum is independent of amplitude of vibration so as long as θ is not large enough for sin θ ≠ θ, the motion is S.H.M. if the amplitude of a simple pendulum increase, the angle θ increases.

Now, sin θ ≠ θ. In this situation the motion of simple pendulum will be oscillatory but not simple harmonic.

Speed of pendulum, \(v = \omega\sqrt{a^2 \,-\,x^2}\)

\(v=\frac{2\pi}{T}\sqrt{a^2-\frac{a^2}{4}}\)

= \(\frac{\sqrt3a\pi}{T}\)

No effect on time period when amplitude of pendulum is increased or decreased.

(b) √v²₁ + v²₂

(b) Parabola

Sound waves are mechanical waves whose velocity

\(v =\sqrt{\frac{\gamma RT}{M}} \)

Light waves are non-mechanical waves or electromagnetic waves for which

c = \(\frac{1}{\sqrt{\mu_0 ε_0}}\) where μ₀ is absolute magnetic permeability of free space and ε₀ is absolute electrical permittivity of free space. Therefore, v depends upon T, but c does not.

R.m.s speed of molecules of a gas

c = \(\sqrt{\frac{3P}{ρ}}\)

c = \(\sqrt{\frac{3RT}{M}}\) (I) [M = Molar mass]

∵ PV = nRT

n = 1

Or P = \(\frac{RT}{V}\) 

∴ \(\frac{p}{ρ}=\frac{RT}{M}\)  [∴\(\frac{P}{δ}=\frac{\frac{RT}{v}}{\frac{M}{v}}=\frac{RT}{M}\) ] 

Speed of sound wave in gas, v = \(\sqrt{\frac{rP}{ρ}}\) 

v = \(\sqrt{\frac{rRT}{M}}\) (II)

Dividing eqn (II) by eq.n (I),

 \(\frac{c}{v}=\frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{rRT}{M}}}\)

 \(\frac{c}{v}=\sqrt{\frac{3}{r}}\) [r = adiabatic constant for diatomic gas]

r = \(\frac{7}{5}\) 

Thus, \(\frac{c}{v}\) = constant

Velocity of sound in CO₂ is less than that in air. Therefore balloon will behave as a convex lens for sound waves. In hydrogen, velocity of sound is greater than that in air. Therefore, balloon filled with hydrogen will behave as a concave lens.

Data : μ = 10 A.m² , Bh = 3.9 × 10^-5 T, θ = 10° 

The magnitude of the torque is τ = – μB_h sin θ 

= (10)(3.9 × 10^-5) sin 10° 

= (3.9 × 10^-4)(0.1736) = 6.770 × 10^-5 N.m

The general expression for the displacement (x) of a particle performing SHM is x = A sin (ωt + α)

(1) Let T be the period of the SHM and x the displacement after a further time interval T. Then

x₁ = A sin [ω(t + T) + α] 

= A sin (ωt + ωT + α) 

= A sin (ωt + α + ωT)

Since T ≠ 0, for x₁ to be equal to x, we must have (ωT)_min = 2π.

Hence, the period (T) of SHM is T = 2π/ω This is the expression for the period in terms of the constant co, the angular frequency.

(2) If m is the mass of the particle and k is the force constant, ω = \(\sqrt{k/m}\).

∴T = \(\frac{2\pi}\omega\) = \(\frac{2\pi}{\sqrt{k/m}}\) = 2π \(\sqrt{\frac mk}\)

(3) The acceleration of a particle performing SHM has a magnitude a = ω²x

∴ ω = \(\sqrt{a/x}\)

= \(\sqrt{acceleration\,per\,unit\,displacement}\)

∴ T = \(\frac{2\pi}\omega\) = \(\frac{2\pi}{\sqrt{acceleration\,per\,unit\,displacement}}\)

Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement. The differential equation of angular SHM is

I\(\frac{d^2\theta}{dt^2}\) + c θ = 0 … (1)

where I = moment of inertia of the where I = moment of inertia of the oscillating body,

\(\frac{d^2\theta}{dt^2}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,

\(\therefore\) \(\frac{d^2\theta}{dt^2}\) + \(\frac cI\theta\) = 0

Let \(\frac cI\) = \(\omega^2\), a constant. Therefore the angular frequency, \(\omega\) = \(\sqrt{\frac{c}{I}}\) and the angular acceleration,

a = \(\frac{d^2\theta}{dt^2}\) = - \(\omega^2\)\(\theta\).....(2)

The minus sign shows that the \(\propto\) and \(\theta\) have opposite directions. The period T of angular SHM is 

T = \(\frac{2\pi}{\omega}\) = \(\frac{2\pi}{\sqrt{\frac{C}{I}}}\) = 2\(\pi\) \(\sqrt{\frac{I}{C}}\).....(3)

This is the expression for the period terms of torque constant. Also from Eq.(2),

\(\omega\) = \(\sqrt{|\frac{x}{\theta}|}\)

= \({\sqrt{angular\,acceleration\,per\,unit\,angular\,displacement}}\) 

\(\therefore\) T =  \(\frac{2\pi}{\omega}\) = \(\frac{2\pi}{\sqrt{|\frac{\propto}{\theta}}|}\) 

= \(\cfrac{2\pi}{\sqrt{angular\,acceleration\,per\,unit\,angular\,displacement}}\)

The total energy of a particle of mass m performing SHM with angular frequency ω, E = 1/2 mω² A²

The maximum acceleration of the particle, a_max = ω² A² 

E = 1/2 mAa_max is the required expression.

Data :  x = 6 sin (3πt +\(\frac{5\pi}6\)) metre Comparing this equation with x = A sin (ωt + α), we get:

1. Amplitude, A = 6 m 

2. ω = 3π rad / s 

∴ Frequency, f = \(\frac{\omega}{2\pi}\) = \(\frac{3\pi}{2\pi}\)= 1.5 Hz 5% 

3. Phase constant, α = \(\frac{5\pi}6\) rad

For a particle of mass m executing SHM with angular frequency ω and amplitude A, its kinetic and potential energies are respectively,

KE = 1/2 mω² (A² – x²) … (1) 

and PE = 1/2 mω²x² … (2) 

Then, the total energy,

E = PE + KE

= 1/2 mω² x² + 1/2 mω² (A² – x²) 

= 1/2 mω² A² …. (3)

Therefore, total energy of the particle is 

1. directly proportional to the mass (E ∝ m), 

2. directly proportional to the square of the amplitude (E ∝ A²) 

3. proportional to the square of the frequency (E ∝f² ), as f = ω/2π

When a body of mass m performs linear SHM, the restoring force on it is always directed towards the mean position and its magnitude is directly proportional to the magnitude of the displacement of the body from the mean position. Thus, if \(\vec F\) is the force acting on the body when its displacement from the mean position is \(\vec x\).

\(\vec F\) = m\(\vec a\) = - kx\(\vec x\)

where the constant k, the force per unit displacement, is called the force constant.

Let k/m = ω² , a constant.

∴ Acceleration, a = –  k/m = - ω²x

∴ The angular frequency,

ω = \(\sqrt\frac km\) = \(\sqrt{|\frac ax|}\)

= \(\sqrt{acceleration\,per\,unit\,displacement}\)

Data : d = 14 cm = 0.14 m

T = 2π \(\sqrt{\frac dg}\) 

= 2(3.142) \(\sqrt{\frac {0.14}{9.8}}\)

= 0.7513 s

Correct option is (B) √3 πnA

(B) displacement is zero

Data : T = 6.28 s, 2A = 20 cm ∴ A = 10 cm, x = 6 cm

v = \(\omega\) \(\sqrt{A^2-x^2}\) = \(\frac{2\pi}T\)\(\sqrt{A^2-x^2}\)

The velocity of the particle at x = 6 cm,

\(\therefore\)v = \(\frac{2\times3.14}{6.28}\)\(\sqrt{(10)^2-(6)^2}\)

= \(\sqrt{100-36}\) = \(\sqrt{64}\) = ± 8 cm/s

Correct option is (C) 0.2 π s

Correct option is (B) π rad

Yes, because the wording of wrist spring wound watch is independent of the ‘g’ but depends upon the P.E. stored in the spring.