Embibe Experts Solutions for Chapter: Simple Harmonic Motion, Exercise 1: Assam CEE 2017

The time period of a simple pendulum inside a stationary lift is $T$. If the lift starts moving upwards with an acceleration of $\frac{g}{3}$, its time period will be, 

A mass $2 \mathrm{kg}$ is suspended from two coupled springs connected in series. The force constants for springs are $24.5 \mathrm{N} {\mathrm{m}}^{-1}$ and $25.5 \mathrm{N} {\mathrm{m}}^{-1}$. The time period of the suspended mass is

An object of mass $4 \mathrm{kg}$ is attached to a spring which is fixed at one end on a rigid support and the mass-spring system is kept on a frictionless table. The object is allowed to execute simple harmonic motion along $X$ - direction. The force constant of the spring is $10 \mathrm{N} {\mathrm{m}}^{-1}$ and the spring is stretched initially a distance of $5 \mathrm{cm}$, the total energy stored in the system is

A particle moves in the $xy$ -plane under the action of superposition of two simple harmonic vibrations. The resultant displacement of the particle is governed by the equation.

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}cos\mathrm{\alpha }=si{n}^2\mathrm{\alpha }$

where $a\mathit{\text{,}}\mathit{ }b$ and $\mathrm{\alpha }$ are positive constants. The particle trajectory $y\left(x\right)$ is linear with a negative slope for

A simple pendulum suspended from the ceiling of a stationary lift has a time period $T_0$. When the lift descends at uniform speed, the time period is $T_1$. When the lift descends with constant acceleration, the time period is $T_2$. Which of the following is correct?