A simple pendulum of length l₁ has a time period of 2 s and another pendulum of length l₂ has a time period of 1 s. Then, the time period of pendulum having length l₁ − l₂ will be

• 2 s

• 1.2 s

• 3 s

• $\sqrt{3}$ s

Springs of spring constants K, 2K,4K, 8K...are connected in series.A mass m kg is attached to the lower end of the last spring and the system is allowed to vibrate. The time period of oscillations is (Given, m= 40 g and k =2.0 N-cm^-1)

• 2.12 s

• 0.126 s

• 1.26 s

• 0.0126 s

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to T_m . If the Young's modulus of the material is Y, then $\frac{1}{\mathrm{Y}}$ is equal to

• $\left[{\left(\frac{{\mathrm{T}}_{\mathrm{m}}}{\mathrm{T}}\right)}^2 - 1\right]\frac{\mathrm{A}}{\mathrm{Mg}}$

• $\left[{\left(\frac{{\mathrm{T}}_{\mathrm{m}}}{\mathrm{T}}\right)}^2 - 1\right]\frac{\mathrm{Mg}}{\mathrm{A}}$

• $\left[1 - {\left(\frac{{\mathrm{T}}_{\mathrm{m}}}{\mathrm{T}}\right)}^2\right]\frac{\mathrm{A}}{\mathrm{Mg}}$

• $\left[1 - {\left(\frac{\mathrm{T}}{{\mathrm{T}}_{\mathrm{m}}}\right)}^2\right]\frac{\mathrm{A}}{\mathrm{Mg}}$

Which of the following graphs represents resonance for low damping ?

• graph (1)

• graph (2)

• graph (3)

• None of these

A magnet is made to oscillate with a particular frequency, passing through a coil as shown in the figure. The time variation of the magnitude of emf generated across the coil during one cycle is

A particle executes simple harmonic motion between x = − A and x = + A. The time taken for it to go from O to A/ 2 is T₁ and to go from A/2 to A is T₂ . Then

• T₁ < T₂

• T₁ > T₂

• T₁ = T₂

• T₁ = 2T₂

A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement y. Which of the following statements is true ?

• TE is zero when x = 0

• PE is maximum when x = 0

• KE is maximum when x = O

• KE is maximum when x is maximum

The KE and PE of a particle executing SHM of amplitude a will be equal when displacement is

• $\frac{\mathrm{a}}{2}$

• $\mathrm{a}\sqrt{2}$

• 2a

• $\frac{\mathrm{a}}{\sqrt{2}}$

The time period of the second's hand of a watch is

• 1 h

• 1 s

• 12 h

• 1 min

A particle starts SHM from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3 $\frac{\mathrm{E}}{4}$ · Its displacement at that instant is

• $\frac{\mathrm{a}}{\sqrt{2}}$

• $\frac{\mathrm{a}}{2}$

• $\frac{\mathrm{a}}{\sqrt{\left({\displaystyle \frac{3}{2}}\right)}}$

• $\frac{\mathrm{a}}{\sqrt{3}}$