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A) 2 s

Explanations:

Amplitude of particle executing SHM 0.2m. Time period  T=24s

Hence

\(x=0.2sin\omega t\)

\(sin\omega t=\frac{1}{2}=sin \frac{\pi}{6}\)

\(\frac{2\pi}{24} t = \frac{\pi}{6}\)

t= 2s

Woofer or base refers to low frequency part of musical sound.

The ratio between the distance traveled by the oscillator in one time period and amplitude is Four.

Forces on the spring are

F₁ = −kx (left spring)

F₂ = −kx (right spring).

Restoring Force, F = F₁ + F₂ = −2kx

∴ F = −2kx towards left

(i) Velocity of the oscillator is zero at extreme positions namely A, C, E and G.

(ii) Speed of oscillator is maximum is at mean position namely B, D, F and H.

Given: amplitude (r) = 5 cm, period(t) = 0.2 sec

To find acceleration and velocity at displacement, '5' cm, '3' cm and '0' cm

(a) v = 2π/t √{(r² - y²)} = 2π/t √{(5)² - (5)²} = 0 ms^-1

Acceleration = (2π/t)²y

= {-4π²}/{0.2 x 0.2} x 5 = -5π² ms^-2

(b) Velocity (v) = 2π/t √{(5)² - (3)²}

= 2π/t √{(25) - (9)} = 2π/t x 4 cms^-1

= 2π/0.2 x 4/100 ms^-1 = 0.4π ms^-1

Acceleration (a) = -(2π/t)²y

= {-4π²}/{0.2 x 0.2} x {3}/{100} ms^-2

= -3π² - ms^-2

(c) Velocity (v) = 2π/t √{(5)² - (0)²}

= 2π/t x 5 = {2π x 5 x 10}/{2 x 100} ms^-1

= π/2 ms^-1 = 0.5 π ms^-1

Acceleration (a) = -(2π/t)² x y

{-4π² x 0}/{t²} = 0 ms^-2

B) 2.22rad/s

EXPLANATION:

ω=√K/M=√4.48/0.98=2.22 rad/sec

Velocity v = ω\(\sqrt{r^2\,-\,y^2}\)

At y = s, let v = v₀ then,

\(v^2_0\) = ω²(r² − s² ) …(i)

Due to blow, the new velocity at y = s is

V = 2v₀, r = r′

So (2v₀)² = ω²(r'² − s² ) …(ii)

Dividing (ii) by (i),

 \(4=\frac{r'^2-s^2}{r^2-s^2}\)

On solving r′= \(\sqrt{4r^2\,-\,3s^2}\)

As the particle starts from rest, it must start from the extreme position. Hence, when t = 0, x = r, where r is the required amplitude.

Using the relation, x = rcos ωt

Or r − x₁ = rcos ω × 1

= r cos ω …(i)

And r − (x₁ + x₂ ) = rcos ω × 2

= rcos 2ω

Or r − x₁ − x₂ = r(2cos²ω − 1) …(ii)

Solving (i) and (ii),

r = \(\frac{2x^2_1}{3x_1-x_2}\)

In steel string, sound will reach the other end earlier, because (y/p) is larger for steel.

Phase difference = \(\frac{\pi}{2}\) = 90º.

1. displacement at t = 0

x(0) = 5cos(2π 0 + π/4)

⇒ x = \(\frac{5}{\sqrt2}\)m

2. Angular frequency 

ω = 2π radian/s

3. Vmax = aω = 2π × 5 = 10π m/s

Newton’s formula for velocity of sound in air:

v = \(\sqrt{\frac{E}{ρ}}\) …(1)

Where v = velocity of sound in the medium.

E = coefficient of elasticity of medium.

ρ = density of medium.

Newton’s formula for velocity of sound in gas: v = \(\sqrt{\frac{k_i}{ρ}}\) 

Since, a gas has only one type of elasticity, i.e., bulk modulus (K),

Sound travel through a gas in the form of compression and rarefactions.

Newton assumed that changes in pressure and volume of a gas, when sound waves are propagated through it, are isothermal. Using coefficient of isothermal elasticity, i.e., K_i in (eqn. 1)

v = \(\sqrt{\frac{K_i}{ρ}}\) … (2)

Error in Newton’s formula:

Let us consider the velocity of sound in air at N.T.P.

v = \(\sqrt{\frac{P}{ρ}}\) … (3)

As, P = hdg

h = 0.76 m of Hg column

d = 13.6 × 10³ kgm^-3 

∴ P = 0.76 × 13.6 × 10^-3 × 9.8 Nm^-2

Density of air, P = 1.293 kg/m³ 

Here from equ (3) = \(\sqrt{\frac{0.76\times13.6\times 10^3\times9.8}{1.293}}\) = 280 ms^-1

The experimental value of the velocity of sound in air at N.T.P. is 332 ms^-1 

Difference between the experimental and theoretical value of velocity or sound in air

= (332 – 280) ms^-1

= 52 ms^-1

Percentage Error = \(\frac{52}{332}\) × 100

= 15.7 %

Or ≈ 16% 

Laplace’s correction: 

According to Laplace, the changes in pressure and volume of a gas, when sound waves are propagated through it, are not isothermal, but adiabatic. This is because.

(i) Velocity of sound in a gas is quite large.

(ii) A gas is a bad conductor of heat.

Using the coefficient of adiabatic elasticity, i.e., K_i instead of K_a :

v = \(\sqrt{\frac{K_i}{ρ}}=\sqrt{\frac{K_a}{ρ}}\) 

Calculation of ‘K_a’

Let P be the initial pressure and V be the initial volume of the certain mass of the gas. Under adiabatic condition PV^r  = constant …(1)

Where γ = \(\frac{c_p}{c_v}\) = ratio of two principal specific heats of the gas.

Differentiating both sides of eqn. 1. P(γ^{\(V^{\gamma-1}\)}dV) + V′ (dP) = 0

or γ^{\(PV^{\gamma-1}\)}dV = −V^γ(dP)

or γ^P = − \(\frac{V^\gamma}{V^{\gamma-1}}(\frac{dP}{dV})\) 

= \(\frac{dP}{\frac{dV}{V}}\)= K_a 

(By definition) 

∴ K_a = γ^P 

Corrected formula: Substituting this value of K_a in

v = \(\sqrt{\frac{K_a}{ρ}}=\sqrt{\frac{\gamma^P}{ρ}}\)

The value of γ depends on nature of the gas.

(b) independent of a

Since V(x) = Kx² , the motion is simple harmonic. In SHM, the time period is independent of the amplitude of oscillation.