PYQ Vault
EbooksTutor

Ratio of distances travelled in the first and second seconds, starting from the mean position

The same question, asked 5 times: 20 Apr 2025 S1, 22 Apr 2025 S2, 19 Apr 2026 S1, 12 May 2026 S1*, 13 May 2026 S1. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    20 Apr 2025 S1

    48
    A particle oscillates in straight line simple harmonically with period 8 second and amplitude 2 m\sqrt{2}\text{ }m. Particle starts from mean position. The ratio of the distance travelled by it in 1st 1^{\text{st~}} second of its motion to that in 2nd 2^{\text{nd~}} second is (sin⁡45∘=1/2,sin⁡π2=1)\left( \sin45^{\circ}= 1/\sqrt{2},\sin\frac{\pi}{2}= 1 \right)

    Tap an option to check your answer.

  2. Question 2

    22 Apr 2025 S2

    24
    A particle performing linear S.H.M. has period 8 seconds. At time t=0t = 0, it is in the mean position. The ratio of the distances travelled by the particle in the 1st 1^{\text{st~}} and 2nd 2^{\text{nd~}} second is (cos⁡45∘=1/2)\left( \cos45^{\circ}= 1/\sqrt{2} \right)

    Tap an option to check your answer.

  3. Question 3

    19 Apr 2026 S1

    118
    A particle executes a simple harmonic motion with a periodic time 88 second. At time t=0t = 0, it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is (sin⁡45∘=cos⁡45∘=12,sin⁡90∘=cos⁡0∘=1)\left(\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1\right)

    Tap an option to check your answer.

  4. Question 4

    12 May 2026 S1

    118
    A particle performing S.H.M., has a period of 1212 s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is {cos⁡(π/6)=32}\{\cos(\pi/6) = \dfrac{\sqrt{3}}{2}\}

    Tap an option to check your answer.

  5. Question 5

    13 May 2026 S1

    118
    A particle oscillates simple harmonically along a straight line with period 88 s and amplitude 828\sqrt{2} m. It starts from mean position then the ratio of the distances travelled by it in the 2nd2^{nd} second and the first second of its motion is (sin⁡45∘=12)\left(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\right) (sin⁡90∘=1)(\sin 90^\circ = 1)

    Tap an option to check your answer.