The time period of a particle in simple harmonic motion is 8 seconds. At t=0 it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is

Question

The time period of a particle in simple harmonic motion is 8 seconds. At t=0 it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is (A) (1/2) (B) (1/√2) (C) (1/√2-1) (D) (1/√3)

Tardigrade Admin · Accepted Answer

Since the particle is at the mean position at

t = 0

, so its displacement

x

any time

t

is

x = A sin (ω t) = A sin (\frac{2 π}{T} t) = A sin (\frac{π}{4} t)

∵ T = 8 s)

The distance travelled in the

I^{s t}

second is

x_{1} = A sin (\frac{π}{4}) = \frac{A}{2}

and that travelled in the

2^{n d}

second is

x_{2} = A sin (\frac{π}{2}) - A sin (\frac{π}{4}) = A - \frac{A}{2} = A (1 - \frac{1}{2})

∴ \frac{x _{1}}{x _{2}} = \frac{\frac{A}{2}}{A ( 1 - \frac{1}{2} )} = \frac{1}{2 - 1}

Q. The time period of a particle in simple harmonic motion is 8 seconds. At $t = 0$ it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is

$\frac{1}{2}$

$\frac{1}{2}$

$\frac{1}{2 - 1}$

$\frac{1}{3}$

Q. The time period of a particle in simple harmonic motion is 8 seconds. At t=0 it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is

21​

2​1​

2​−11​

3​1​

Q. The time period of a particle in simple harmonic motion is 8 seconds. At $t = 0$ it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is

$\frac{1}{2}$

$\frac{1}{2}$

$\frac{1}{2 - 1}$

$\frac{1}{3}$