Q.37. A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec. t

Q: What is the correct answer for JEE Main Physics Q37?

The correct answer is C. Given: Amplitude = A Time period T = 5 s The displacement of a particle in SHM starting from the extreme position ($x=A$ at $t=0$) is given by: $x(t) = A \cos(\omega t)$ where $\omega = \frac{2\pi}{T}$ is the angular frequency. We need to find the time $t$ when the particle moves from $x=A$ to $x=A/\sqrt{2}$. Substitute $x = A/\sqrt{2}$ into the equation: $A/\sqrt{2} = A \cos(\omega t)$ $\frac{

Q: Q.37. A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec. the time required by it to move from to is ____ sec.

The correct answer is C. Given: Amplitude = A Time period T = 5 s The displacement of a particle in SHM starting from the extreme position ($x=A$ at $t=0$) is given by: $x(t) = A \cos(\omega t)$ where $\omega = \frac{2\pi}{T}$ is the angular frequency. We need to find the time $t$ when the particle moves from $x=A$ to $x=A/\sqrt{2}$. Substitute $x = A/\sqrt{2}$ into the equation: $A/\sqrt{2} = A \cos(\omega t)$ $\frac{

Q: What are the options for this JEE Main Physics question?

A. 1/4 · B. 5/4 · C. 5/8 · D. 3/8

Question

Q.37. A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec. the time required by it to move from  to  is ____ sec.

FirstInTest · Accepted Answer

Given:
Amplitude = A
Time period T = 5 s

The displacement of a particle in SHM starting from the extreme position ($x=A$ at $t=0$) is given by:
$x(t) = A \cos(\omega t)$
where $\omega = \frac{2\pi}{T}$ is the angular frequency.

We need to find the time $t$ when the particle moves from $x=A$ to $x=A/\sqrt{2}$.

Substitute $x = A/\sqrt{2}$ into the equation:
$A/\sqrt{2} = A \cos(\omega t)$
$\frac{1}{\sqrt{2}} = \cos(\omega t)$

The principal value for which $\cos(	heta) = \frac{1}{\sqrt{2}}$ is $	heta = \frac{\pi}{4}$.
So, $\omega t = \frac{\pi}{4}$.

Substitute $\omega = \frac{2\pi}{T}$:
$\frac{2\pi}{T} t = \frac{\pi}{4}$

Solve for t:
$t = \frac{\pi}{4} 	imes \frac{T}{2\pi} = \frac{T}{8}$

Given $T = 5$ s:
$t = \frac{5}{8}$ s.

Answer

1/4

Answer

5/4

Answer

5/8

Answer

3/8

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