## Question

### Solution

Correct option is

196Hz

Let v1 be the frequency of the wire when its vibrating length is

L1 = 48 cm and v2 when L2 = 50 cm. Since  If v is the frequency of the tuning fork, then

Which gives v = 196Hz.

#### SIMILAR QUESTIONS

Q1

Two springs A and B have force constants k1 and k2 respectively. The ratio of the work done on A to that done on B in increasing their lengths by the same amount is

Q2

Two springs A and B have force constants k1 and k2 respectively. The ratio of the work done on A to that done on B when they are stretched by the same force is

Q3

The mass m shown in figure is displaced vertically by a small amount and released. If the system oscillates with a period of 2 seconds, the value of the spring constant k is

Q4

A spring of force constant k is cut into three equal pieces. If these three pieces are connected in parallel, the force constant of the combination will be

Q5

Two springs of force constants k1 and k2 are connected as shown in figure. The time period of vertical oscillations of mass m is given by

Q6

A sonometer wire of length 120 cm is divided into three segments of lengths in the ratio of 1 : 2 : 3. What is the ratio of their fundamental frequencies?

Q7

A particle is in linear simple harmonic motion between two extreme pointsA and B, 10 cm apart (see fig). If the direction from A to B is taken as positive direction, what are signs of displacement x, velocity V and acceleration a, when the particle is at A

Q8

A simple harmonic motion of amplitude A has a time period T. The acceleration of the oscillator when its displacement is half the amplitude is

Q9

A body oscillates harmonically with amplitude 0.05 m. At a certain instant of time its displacement is 0.01 m and acceleration is 1.0 ms-2. What is the period of oscillation?

Q10

The displacement x (in centimetres) of an oscillating particle varies with time t (in seconds) as

The magnitude of the maximum acceleration of the particle is