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Simple Harmonic Motion

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New answer posted

2 months ago

Simple Harmonic Motion Class 11th

The equation y = acos²ωt + asin²ωt is

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alok kumar singh

Contributor-Level 10

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

Two simple harmonic motion are represented by,

And

The ratio of their amplitudes is

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Raj Pandey

Contributor-Level 9

$A_{1} = \sqrt[]{{(5)}^{2} + {(5 \sqrt[]{3})}^{2}} = 10$

$A_{2} = 5$

Then, $\frac{A_{1}}{A_{2}} = \frac{10}{5} = \frac{2}{1}$

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

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alok kumar singh

Contributor-Level 10

$y = A \sin ? \frac{2 π t}{T}$

$t_{2} - t_{1} = \frac{T}{2 π} [\sin^{- 1} ? \frac{x_{1}}{A} - \sin^{- 1} ? \frac{x_{2}}{A}]$

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

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Vishal Baghel

Contributor-Level 10

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

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Vishal Baghel

Contributor-Level 10

y = A sin (2πt/T)
t? - t? = (T/2π) [sin? ¹ (x? /A) - sin? ¹ (x? /A)]

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

A body is executing simple harmonic motion with frequency ' $n$ ', the frequency of its potential energy is:

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alok kumar singh

Contributor-Level 10

Displacement equation of SHM of frequency ' $n$ '

$x = A s i n ? (ω t) = A s i n ? (2 π n t)$

Now,

Potential energy $U = \frac{1}{2} {k x}^{2} = \frac{1}{2} {K A}^{2} {s i n}^{2} ? 2 π n t$

$= \frac{1}{2} {k A}^{2} [\frac{1 - c o s ? 2 π 2 n t}{2}]$

So frequency of potential energy $= 2 n$

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1/a s. The value of 'a' to the nearest integer is-----------.

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Raj Pandey

Contributor-Level 9

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

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New answer posted

3 months ago

Simple Harmonic Motion Class 11th

Two particles A and B of equal masses are suspended from two massless springs of spring constants K₁ and K₂ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

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alok kumar singh

Contributor-Level 10

A? √* (K? /m)* = A? √* (K? /m)* ⇒ A? √K? = A? √K?

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New question posted

4 months ago

Simple Harmonic Motion Class 11th

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin(ωt + φ). If the position and velocity of the particle at t = 0s are 2cm and 2ω cm s?¹ respectively, then its amplitude is x√2 cm where the value of x is _____.

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New answer posted

4 months ago

Simple Harmonic Motion Class 11th

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

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