A simple pendulum has time period’t’. Its time period in a lift which is moving upwards with acceleration 3 ms^–2 is

\(t\sqrt{\frac{9.8}{12.8}}\)
\(t\sqrt{\frac{12.8}{9.8}}\)
\(t\sqrt{\frac{9.8}{6.8}}\)
\(t\sqrt{\frac{6.8}{9.8}}\)

Solution

y = 2 (cm) sin [^πt⁄₂ + φ] what is the maximum acceleration of the particle doing the S.H.M.

^π⁄₂ cm/s²
^π²⁄₂ cm/s²
^π²⁄₄ cm/s²
^π⁄₄ cm/s²

Solution

A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium. It undergoes oscillations with a time period T then

T α m
T α ρ
T α 1/A
T α 1/ρ

Solution

A rectangular block of mass m and area of cross-section

A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium. It undergoes oscillations with a time period T then: T α m

The displacement of a S.H.M. doing particle when K.E. = P.E. (amplitude =4 cm) is

2√2 cm
2 cm
¹⁄_√2 cm
√2 cm

Solution

The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface is

1/6 m
6 m
1/36 m
36 m

Solution

A particle executes S.H.M.having time period T, then the time period with which the potential energy changes is

Solution

P.E. changes from zero to maximum twice in each vibration so its time period is T/2

The potential energy of a particle (U_x) executing S.H.M. is given by

U_x = ^k⁄₂ (x - a)²
U_x = k₁x + k₂x² + k₃x³
U_x = Ae^-bx
U_x = a constant

Solution

A mass M is suspended from a spring of negligible mass.The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is in creased by m, the time period becomes ^5T⁄₃. Then the ratio of ^m⁄_M is