Question

The equation of a simple harmonic progressive wave is given by y = 4 sin π ( $\frac{t}{0.02} - \frac{x}{75}$ ) cm. Find the displacement and velocity of the particle at a distance of 50 cm from the origin and at the instant 0.1 second (all quantities are in c.g.s. units) (Ans : Displacement = 3.464 cm, Velocity of the particle = 3.14 m/s)

Answer 1

displacement of particle , y = 4sinπ(t/0.02 - x/75)
here t = 0.1 sec and x = 50cm

so, y = 4sinπ(0.1/0.02 - 50/75)

y = 4sinπ(5 - 2/3)

y = 4sinπ(13/3) = 4sin(13π/3)

y = 4sin(4π + π/3) = 4sinπ/3 = 4 × √3/2

y = 2√3 = 2 × 1.732 = 3.464 cm


y = 4sinπ(t/0.02 - x/75)

differentiate with respect to t,

dy/dt = (4/0.02)πcosπ(t/0.02 - x/75)

v = 200πcosπ(t/0.02 - x/75)

at t = 0.1 sec and x = 50cm

v = 200πcosπ(0.1/0.02 - 50/75) = 200πcos(13π/3)

v = 200πcos(4π + π/3) = 200πcosπ/3

v = 200π × 1/2 = 100π = 314 cm/s or 3.14 m/s

Answer 2

Answer:

Explanation:

Given :

T = 435.5 dyne/cm = 0.4355 N/m,

θ = 14

0

ρ = 13600 kg/m³  

d = 1 cm

∴ r = 0.5 cm

= 5 × 10⁻³ m  

T = rh ρ g/ 2 cos θ  

h = 2T cos θ/ r g ρ  

∴ h = 2 x0.4355x cos140⁰/ 5 x10⁻³x 13600 x9.8

= 0.8710x cos( 90 + 50  )/ 5x 13.6 x9.8  

=0.8710( –sin 50  ) /5x 13.6 x9.8

=–0.8710x 0.7660/ 68.0x 9.8  

= – 1.001 × 10⁻³m  

∴ h = – 1.001 mm

here  Negative sign indicates that mercury level will be lowered by 1.001 mm.

Hence to get correct reading h = 1.001 mm has to added.  

∴ h = 1.001 mm

1.HNO3+H2SO4

2.no reagent just Br3

3.zinc dust

4.CCl4+HCL