Question

a particle is executing from simple harmonic motion. velocity of particle at the displacement 4cm and 6 cm from the mean position s are 3cm/s and 2 cm/s respectively. Find its time period and amplitude​

Answer 1

The question is meant to give you a practice of the application of the formula of velocity in terms of displacement. If you are aware of the formula : v = ω√(A² - x²) then this question is merely all about calculation for you,else it's a skip to next kind question xD

So,plug the given data in the formula mentioned.

Given data :

• x1 = 4 cm,v1 = 3 cm/s
• x2 = 6 cm,v2 = 2 cm/s

To find :

• Amplitude (A) and time period (T)

Solution :

First write the equation using the x1 and v1 value,

v1 = ω√(A² - x1²)

Square both sides to simplify a bit,

v1² = ω² (A² - x1²)

Put the values,

3² = ω² (A² - 4²)

9 = ω² (A² - 16) -----> (1)

Now write a new equation using v2 and x2,

v2 = ω√(A² - x2²)

Squaring again,

v2² = ω²(A² - x2²)

Put the values,

2² = ω² (A² - 6²)

4 = ω² (A² - 36) -----> (2)

Next divide equation (1) by (2),

9/4 = ω²(A² - 16)/ω²(A² - 36)

9/4 = A² - 16/ A² - 36

Cross multiply,

9 (A² - 36) = 4 (A² - 16)

9A² - 324 = 4A² - 64

9A² - 4A² = 260

5A² = 260

A² = 260/5

A² = 52

A = √52

A = 7.2

A = 7 (approx)

Now substitute the obtained value of A in any of the two equation,i am using equation (1),

9 = ω² (7² - 16)

9 = ω² (49 - 16)

9 = ω² (33)

9/33 = ω²

3/11 = ω²

0.27 = ω²

ω = √0.27

ω = 0.51

Hoping you know the formula for time period,T = 2π/ω

Just put the value of ω in above formula,

T = 2π/0.51

T = 12.31

T = 12

Therefore, amplitude (A) of oscillation is 7 cm and time period T of the oscillation is 12 second.