Question

When the displacement in S.H.M. is 1/3rd of the amplitude. What fraction of total energy is kinetic energy and what fraction is potential energy ?

Answer 1

Let amplitude of SHM is A.
Then, displacement of SHM,x = A/3 [ from question]
total energy = 1/2 m$\omega^2$ A²

Kinetic energy, 1/2 m$\omega^2$(A² - x²)

= 1/2 m$\omega^2$(A² - A²/9)

= 1/2 m$\omega^2$ 8A²/9

= 1/2(8/9) m$\omega^2$A² ....(2)

Fractional of total energy to kinetic energy = 1/2m$\omega^2$A²/1/2(8/9)m$\omega^2$A²

= 9/8 or 9 : 8


Potential energy = 1/2 m$\omega^2$x²

= 1/2m$\omega^2$A²/9

= 1/2(1/9) m$\omega^2$A²

fractional of total energy to potential energy = 1/2m$\omega^2$A²/1/2(1/9)m$\omega^2$A²

= 9 : 1

Answer 2

Answer:

time period of pendulum , T = 1.44s

Answer:

r = 0.5mm = 0.5 × 10⁻³m

= 5 × 10⁻⁴ m

T = 0.04 N/m

ρ = 0.8 gm/cc

= 0.8x 10 ⁻³/10 ⁻⁶  = 800 kg/m3  

θ = 10⁰

g = 9.8 m/s²

T = rh ρ g /2 cos θ  

∴ h = 2T cos θ/ r g ρ

∴ h = 2x0.04 cos10/ 5 x10⁻⁴x 800x 9.8  

h = 2x 0.04x 0.9848/ 5x 800x 9.8 10⁻⁴

h = 2.01 × 10⁻² m

∴ The height of the capillary rise is 2.01 × 10⁻² m

Use formula,

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

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

Physics • 19 hours ago

Answer:

Given :

R = 0.5cm = 0.5 × 10⁻²m

n = No. of drops formed = 10⁶

ρ = 13600 kg/m3 ,  

T = 0.465 N/m  

W = T∆A  

P.E = mgh

Volume of single big drop

V = 4 / 3 π R ³

Volume of single small droplet = 4 /3  π r³

Volume of n droplets = n × 4/ 3  πr³

∴ 4/ 3  πR³ = n 4/ 3  π r³

∴ R³ = nr³

∴ r = R /∛n = 0.5x 10⁻² /∛ 10⁶

∴ r = 0.5 × 10⁻²m  /10²

r=0.5 x 10⁻⁴m

The single drop is fallen from height h, hence its P.E. = mgh

But, P.E = Work done due to change in area ...(i)

Change in surface area ∆A = (n4πr²2 – 4πR²)

Also, W = T∆A =

T (n4πr ² – 4πR ² )  

W = 4πT (nr² – R² )

∴ eq., (i) becomes,

mgh = 4πT (nr²– R² )

But, m = 4 /3 πR³ρ

∴ 4 /3 πR³ρgh = 4πT (nr² – R² )

R³ gh   ρ/3 = T (nr² – R²)  

∴ h = 3T (nr² – R²  )/ R ³ ρ g

=3x0.465[10 ⁶ (0.5x10⁻⁴)² – (0.5x10⁻²)²]/ (0.5x10⁻²)3x13600x9.8

H=3x0.465[0.25x10⁻²)- (0.25x10⁻⁴)]/0.125x1.36x10⁴ x 9.8

On solving

=3x0.465x25x0.09/12.5x1.36x9.8

H=0.2072m

$.$

Squaring both sides,

T² = 4π²l/g

Physics • 19 hours ago

g = 4π²l/T²

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

= 4(3.14)² × 0.51/(1.44)²

= (3.142)² × 0.51/(0.36 × 1.44)

Physics • 19 hours ago

Answer:

Given :

d1 = 1mm  

∴ r 1 = d 1/2 = 1/ 2

= 0.5 mm = 0.05 cm

d2 = 1.1 mm

∴ r 2 = d2/ 2 = 1.1/ 2 = 0.55 mm = 0.55 cm

T = 75 dyne/cm

F = Tl  

l = 2π (r1 + r2 )

F = Tl  

= T × 2π (r1 + r2 )  

∴ F = 2πT (r1 + r2 )

= 2 × 3.142 × 75 (0.05 + 0.055)

= 2 × 3.142 × 75 × 0.105  

∴ F = 150 × 3.142 × 0.105

∴ F = 49.49 dynes

$.$

≈ 9.712 m/s²

Hence, experimental value of g = 9.712m/s²

Physics • 19 hours ago

Answer:

given :

nD = tuning fork D frequency = 340 Hz

nC =  tuning fork C  frequency=8 – 4 = 4 beats per second.

First nC ± nD = 8 (before filing)

nC ± nD = 4 (after filing)

From given condition nC ± nD = 8  

∴ nC ± 340 = 8

∴ nC = 340 + 8 = 348 Hz

or nC = 340 – 8 = 332 Hz

when tuning fork C is filed then nC ± nD = 4  

∴ nC ± 340 = 4  

∴ nC = 340 + 4 = 344 Hz  

or nC = 340 – 4 = 336 Hz

The frequency of tuning fork increases on filing. Hence nC ≠ 344 Hz.  

If original frequency of tuning fork C is taken 332 Hz, then on filing both the value 344 Hz, and 336 Hz are greater. Also it produces 4 beats per second with tunning fork D.

∴ frequency of tuning fork C = 332 Hz

∴ nC = 332 Hz.

abhi178 • ★ Brainly Teacher ★

Physics • 19 hours ago

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